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Preamble: The following subproblems refer to the exponential function $e^{-t / 2} \cos (3 t)$, which we will assume is a solution of the differential equation $m \ddot{x}+b \dot{x}+k x=0$.
What is $b$ in terms of $m$? Write $b$ as a constant times a function of $m$. | We can write $e^{-t / 2} \cos (3 t)=\operatorname{Re} e^{(-1 / 2 \pm 3 i) t}$, so $p(s)=m s^{2}+b s+k$ has solutions $-\frac{1}{2} \pm 3 i$. This means $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$. Then $b=\boxed{m}$, | Differential Equations (18.03 Spring 2010) | 100 |
Preamble: The following subproblems refer to the differential equation. $\ddot{x}+4 x=\sin (3 t)$
Subproblem 0: Find $A$ so that $A \sin (3 t)$ is a solution of $\ddot{x}+4 x=\sin (3 t)$.
Solution: We can find this by brute force. If $x=A \sin (3 t)$, then $\ddot{x}=-9 A \sin (3 t)$, so $\ddot{x}+4 x=-5 A \sin (3 t)$. Therefore, when $A=\boxed{-0.2}, x_{p}(t)=-\sin (3 t) / 5$ is a solution of the given equation.
Final answer: The final answer is -0.2. I hope it is correct.
Subproblem 1: What is the general solution, in the form $f_0(t) + C_1f_1(t) + C_2f_2(t)$, where $C_1, C_2$ denote arbitrary constants? | To find the general solution, we add to $x_{p}$ the general solution to the homogeneous equation $\ddot{x}+4 x=0$. The characteristic polynomial is $p(s)=s^{2}+4$, with roots $\pm 2 i$, so the general solution to $\ddot{x}+4 x=0$ is $C_{1} \sin (2 t)+C_{2} \cos (2 t)$. Therefore, the general solution to $\ddot{x}+4 x=\sin (3 t)$ is given by $\boxed{-\sin (3 t) / 5+ C_{1} \sin (2 t)+C_{2} \cos (2 t)}$. | Differential Equations (18.03 Spring 2010) | 101 |
What is the smallest possible positive $k$ such that all functions $x(t)=A \cos (\omega t-\phi)$---where $\phi$ is an odd multiple of $k$---satisfy $x(0)=0$? \\ | $x(0)=A \cos \phi$. When $A=0$, then $x(t)=0$ for every $t$; when $A \neq 0$, $x(0)=0$ implies $\cos \phi=0$, and hence $\phi$ can be any odd multiple of $\pi / 2$, i.e., $\phi=\pm \pi / 2, \pm 3 \pi / 2, \pm 5 \pi / 2, \ldots$ this means $k=\boxed{\frac{\pi}{2}}$ | Differential Equations (18.03 Spring 2010) | 102 |
Preamble: The following subproblems refer to the differential equation $\ddot{x}+b \dot{x}+x=0$.\\
What is the characteristic polynomial $p(s)$ of $\ddot{x}+b \dot{x}+x=0$? | The characteristic polynomial is $p(s)=\boxed{s^{2}+b s+1}$. | Differential Equations (18.03 Spring 2010) | 103 |
Preamble: The following subproblems refer to the exponential function $e^{-t / 2} \cos (3 t)$, which we will assume is a solution of the differential equation $m \ddot{x}+b \dot{x}+k x=0$.
Subproblem 0: What is $b$ in terms of $m$? Write $b$ as a constant times a function of $m$.
Solution: We can write $e^{-t / 2} \cos (3 t)=\operatorname{Re} e^{(-1 / 2 \pm 3 i) t}$, so $p(s)=m s^{2}+b s+k$ has solutions $-\frac{1}{2} \pm 3 i$. This means $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$. Then $b=\boxed{m}$,
Final answer: The final answer is m. I hope it is correct.
Subproblem 1: What is $k$ in terms of $m$? Write $k$ as a constant times a function of $m$. | Having found that $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$ in the previous subproblem, $k=\boxed{\frac{37}{4} m}$. | Differential Equations (18.03 Spring 2010) | 104 |
Preamble: In the following problems, take $a = \ln 2$ and $b = \pi / 3$.
Subproblem 0: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers.
Solution: Using Euler's formula, we find that the answer is $\boxed{1+\sqrt{3} i}$.
Final answer: The final answer is 1+\sqrt{3} i. I hope it is correct.
Subproblem 1: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{2(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.
Solution: $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-2+2 \sqrt{3} i}$.
Final answer: The final answer is -2+2 \sqrt{3} i. I hope it is correct.
Subproblem 2: Rewrite $e^{3(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.
Solution: $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-8}$.
Final answer: The final answer is -8. I hope it is correct.
Subproblem 3: Rewrite $e^{4(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers. | $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-8-8 \sqrt{3} i}$. | Differential Equations (18.03 Spring 2010) | 105 |
Rewrite the function $\operatorname{Re} \frac{e^{i t}}{2+2 i}$ in the form $A \cos (\omega t-\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$. | $e^{i t}=\cos (t)+i \sin (t)$, and $\frac{1}{2+2 i}=\frac{1-i}{4}$. the real part is then $\frac{1}{4} \cos (t)+$ $\frac{1}{4} \sin (t)$. The right triangle here has hypotenuse $\frac{\sqrt{2}}{4}$ and argument $\pi / 4$, so $f(t)=\boxed{\frac{\sqrt{2}}{4} \cos (t-\pi / 4)}$. | Differential Equations (18.03 Spring 2010) | 106 |
Preamble: The following subproblems refer to the differential equation $\ddot{x}+b \dot{x}+x=0$.\\
Subproblem 0: What is the characteristic polynomial $p(s)$ of $\ddot{x}+b \dot{x}+x=0$?
Solution: The characteristic polynomial is $p(s)=\boxed{s^{2}+b s+1}$.
Final answer: The final answer is s^{2}+b s+1. I hope it is correct.
Subproblem 1: For what value of $b$ does $\ddot{x}+b \dot{x}+x=0$ exhibit critical damping? | To exhibit critical damping, the characteristic polynomial $s^{2}+b s+1$ must be a square, i.e., $(s-k)^{2}$ for some $k$. Multiplying and comparing yields $-2 k=b$ and $k^{2}=1$. Therefore, $b$ could be either one of $=-2, 2$. When $b=-2, e^{t}$ is a solution, and it exhibits exponential growth instead of damping, so we reject that value of $b$. Therefore, the value of $b$ for which $\ddot{x}+b \dot{x}+x=0$ exhibits critical damping is $b=\boxed{2}$ | Differential Equations (18.03 Spring 2010) | 107 |
Find the general (complex-valued) solution of the differential equation $\dot{z}+2 z=e^{2 i t}$, using $C$ to stand for any complex-valued integration constants which may arise. | Using integrating factors, we get $e^{2 t} z=e^{(2+2 i) t} /(2+2 i)+C$, or $z=\boxed{\frac{e^{2 i t}}{(2+2 i)}+C e^{-2 t}}$, where $C$ is any complex number. | Differential Equations (18.03 Spring 2010) | 108 |
Preamble: Consider the first-order system
\[
\tau \dot{y}+y=u
\]
driven with a unit step from zero initial conditions. The input to this system is \(u\) and the output is \(y\).
Derive and expression for the settling time \(t_{s}\), where the settling is to within an error \(\pm \Delta\) from the final value of 1. | Rise and Settling Times. We are given the first-order transfer function
\[
H(s)=\frac{1}{\tau s+1}
\]
The response to a unit step with zero initial conditions will be \(y(t)=1-e^{-t / \tau}\). To determine the amount of time it take \(y\) to settle to within \(\Delta\) of its final value, we want to find the time \(t_{s}\) such that \(y\left(t_{s}\right)=1-\Delta\). Thus, we obtain
\[
\begin{aligned}
&\Delta=e^{-t_{s} / \tau} \\
&t_{s}=\boxed{-\tau \ln \Delta}
\end{aligned}
\] | Dynamics and Control (2.003 Spring 2005) | 109 |
Preamble: Consider the first-order system
\[
\tau \dot{y}+y=u
\]
driven with a unit step from zero initial conditions. The input to this system is \(u\) and the output is \(y\).
Subproblem 0: Derive and expression for the settling time \(t_{s}\), where the settling is to within an error \(\pm \Delta\) from the final value of 1.
Solution: Rise and Settling Times. We are given the first-order transfer function
\[
H(s)=\frac{1}{\tau s+1}
\]
The response to a unit step with zero initial conditions will be \(y(t)=1-e^{-t / \tau}\). To determine the amount of time it take \(y\) to settle to within \(\Delta\) of its final value, we want to find the time \(t_{s}\) such that \(y\left(t_{s}\right)=1-\Delta\). Thus, we obtain
\[
\begin{aligned}
&\Delta=e^{-t_{s} / \tau} \\
&t_{s}=\boxed{-\tau \ln \Delta}
\end{aligned}
\]
Final answer: The final answer is -\tau \ln \Delta. I hope it is correct.
Subproblem 1: Derive an expression for the \(10-90 \%\) rise time \(t_{r}\) in terms of $\tau$. | The \(10-90 \%\) rise time \(t_{r}\) may be thought of as the difference between the \(90 \%\) settling time \((\Delta=0.1)\) and the \(10 \%\) settling time \((\Delta=0.9)\).
\[
t_{r}=t_{\Delta=0.1}-t_{\Delta=0.9}
\]
Therefore, we find \(t_{r}=\boxed{2.2 \tau}\). | Dynamics and Control (2.003 Spring 2005) | 110 |
Preamble: For each of the functions $y(t)$, find the Laplace Transform $Y(s)$ :
$y(t)=e^{-a t}$ | This function is one of the most widely used in dynamic systems, so we memorize its transform!
\[
Y(s)=\boxed{\frac{1}{s+a}}
\] | Dynamics and Control (2.003 Spring 2005) | 111 |
Preamble: For each Laplace Transform \(Y(s)\), find the function \(y(t)\) :
Subproblem 0: \[
Y(s)=\boxed{\frac{1}{(s+a)(s+b)}}
\]
Solution: We can simplify with partial fractions:
\[
Y(s)=\frac{1}{(s+a)(s+b)}=\frac{C}{s+a}+\frac{D}{s+b}
\]
find the constants \(C\) and \(D\) by setting \(s=-a\) and \(s=-b\)
\[
\begin{aligned}
\frac{1}{(s+a)(s+b)} &=\frac{C}{s+a}+\frac{D}{s+b} \\
1 &=C(s+b)+D(s+a) \\
C &=\frac{1}{b-a} \\
D &=\frac{1}{a-b}
\end{aligned}
\]
therefore
\[
Y(s)=\frac{1}{b-a} \frac{1}{s+a}-\frac{1}{b-a} \frac{1}{s+b}
\]
By looking up the inverse Laplace Transform of \(\frac{1}{s+b}\), we find the total solution \(y(t)\)
\[
y(t)=\boxed{\frac{1}{b-a}\left(e^{-a t}-e^{-b t}\right)}
\]
Final answer: The final answer is \frac{1}{b-a}\left(e^{-a t}-e^{-b t}\right). I hope it is correct.
Subproblem 1: \[
Y(s)=\frac{s}{\frac{s^{2}}{\omega_{n}^{2}}+\frac{2 \zeta}{\omega_{n}} s+1}
\]
You may assume that $\zeta < 1$. | First, note that the transform is
\[
\begin{aligned}
Y(s) &=\frac{s}{\frac{s^{2}}{\omega_{n}^{2}}+\frac{2 \zeta}{\omega_{n}} s+1} \\
&=s \cdot \frac{\omega_{n}^{2}}{s^{2}+2 \zeta \omega_{n} s+\omega_{n}^{2}}
\end{aligned}
\]
We will solve this problem using the property
\[
\frac{d f}{d t}=s F(s)-f(0)
\]
therefore
\[
\begin{aligned}
y(t) &=\frac{d}{d t}\left(\frac{\omega_{n}}{\sqrt{1-\zeta^{2}}} e^{-\zeta \omega_{n} t} \sin \left(\omega_{n} \sqrt{1-\zeta^{2}} t\right)\right) \\
&=\boxed{\omega_{n}^{2} e^{-\zeta \omega_{n} t} \cos \left(\omega_{n} \sqrt{1-\zeta^{2}} t\right)-\frac{\zeta \omega_{n}^{2}}{\sqrt{1-\zeta^{2}}} e^{-\zeta \omega_{n} t} \sin \left(\omega_{n} \sqrt{1-\zeta^{2}} t\right)}
\end{aligned}
\]
remember that for this form to be correct, \(\zeta\) must be less than 1 . | Dynamics and Control (2.003 Spring 2005) | 112 |
A signal \(x(t)\) is given by
\[
x(t)=\left(e^{-t}-e^{-1}\right)\left(u_{s}(t)-u_{s}(t-1)\right)
\]
Calculate its Laplace transform \(X(s)\). Make sure to clearly show the steps in your calculation. | Simplify the expression in to a sum of terms,
\[
x(t)=e^{-t} u_{s}(t)-e^{-1} u_{s}(t)-e^{-t} u_{s}(t-1)+e^{-1} u_{s}(t-1)
\]
Now take the Laplace transform of the first, second and fourth terms,
\[
X(s)=\frac{1}{s+1}-\frac{e^{-1}}{s}-\mathcal{L} e^{-t} u_{s}(t-1)+\frac{e^{-1} e^{-s}}{s}
\]
The third term requires some massaging to get it in a form available on the table. The term can be modified into the form of a time delay, by factoring out \(e^{-1}\).
\[
\mathcal{L}\left\{e^{-t} u_{s}(t-1)\right\}=e^{-1} \mathcal{L}\left\{e^{-(t-1)} u_{s}(t-1)\right\}
\]
Now applying the Laplace Transform for a time delay from the table
\[
e^{-1} \mathcal{L}\left\{e^{-(t-1)} u_{s}(t-1)\right\}=\frac{e^{-1} e^{-s}}{s+1}
\]
Substituting this piece back into the expression above gives the solution
\[
X(s)=\boxed{\frac{1}{s+1}-\frac{e^{-1}}{s}-\frac{e^{-1} e^{-s}}{s+1}+\frac{e^{-1} e^{-s}}{s}}
\] | Dynamics and Control (2.003 Spring 2005) | 113 |
Preamble: You are given an equation of motion of the form:
\[
\dot{y}+5 y=10 u
\]
Subproblem 0: What is the time constant for this system?
Solution: We find the homogenous solution, solving:
\[
\dot{y}+5 y=0
\]
by trying a solution of the form $y=A \cdot e^{s, t}$.
Calculation:
\[
\dot{y}=A \cdot s \cdot e^{s \cdot t} \mid \Rightarrow A \cdot s \cdot e^{s t}+5 A \cdot e^{s t}=0
\]
yields that $s=-5$, meaning the solution is $y=A \cdot e^{-5 \cdot t}=A \cdot e^{-t / \tau}$, meaning $\tau = \boxed{0.2}$.
Final answer: The final answer is 0.2. I hope it is correct.
Subproblem 1: If \(u=10\), what is the final or steady-state value for \(y(t)\)? | Steady state implies $\dot{y} = 0$, so in the case when $u=10$, we get $y=\boxed{20}$. | Dynamics and Control (2.003 Spring 2005) | 114 |
A signal \(w(t)\) is defined as
\[
w(t)=u_{s}(t)-u_{s}(t-T)
\]
where \(T\) is a fixed time in seconds and \(u_{s}(t)\) is the unit step. Compute the Laplace transform \(W(s)\) of \(w(t)\). Show your work. | The Laplace Transform of \(x(t)\) is defined as
\[
\mathcal{L}[x(t)]=X(s)=\int_{0}^{\infty} x(t) e^{-s t} d t
\]
therefore
\[
\begin{aligned}
W(s) &=\int_{0}^{\infty} e^{-s t} d t-\left(\int_{0}^{T} 0 d t+\int_{T}^{\infty} e^{-s t} d t\right) \\
&=-\left.\frac{1}{s} e^{-s t}\right|_{0} ^{\infty}-\left(0+-\left.\frac{1}{s} e^{-s t}\right|_{T} ^{\infty}\right) \\
&=\boxed{\frac{1}{s}-\frac{1}{s} e^{-s T}}
\end{aligned}
\] | Dynamics and Control (2.003 Spring 2005) | 115 |
Preamble: Assume that we apply a unit step in force separately to a mass \(m\), a dashpot \(c\), and a spring \(k\). The mass moves in inertial space. The spring and dashpot have one end connected to inertial space (reference velocity \(=0\) ), and the force is applied to the other end. Assume zero initial velocity and position for the elements.
Recall that the unit step function \(u_{S}(t)\) is defined as \(u_{S}(t)=0 ; t<0\) and \(u_{S}(t)=1 ; t \geq 0\). We will also find it useful to introduce the unit impulse function \(\delta(t)\) which can be defined via
\[
u_{S}(t)=\int_{-\infty}^{t} \delta(\tau) d \tau
\]
This means that we can also view the unit impulse as the derivative of the unit step:
\[
\delta(t)=\frac{d u_{S}(t)}{d t}
\]
Solve for the resulting velocity of the mass. | \[
\begin{aligned}
m \ddot{x}_{m} &=u_{s}(t) \\
\dot{x}_{m}=v_{m} &=\int_{-\infty}^{t} \frac{1}{m} u_{s}(t) d t=\boxed{\frac{1}{m} t} \\
\end{aligned}
\] | Dynamics and Control (2.003 Spring 2005) | 116 |
Preamble: For each of the functions $y(t)$, find the Laplace Transform $Y(s)$ :
Subproblem 0: $y(t)=e^{-a t}$
Solution: This function is one of the most widely used in dynamic systems, so we memorize its transform!
\[
Y(s)=\boxed{\frac{1}{s+a}}
\]
Final answer: The final answer is \frac{1}{s+a}. I hope it is correct.
Subproblem 1: $y(t)=e^{-\sigma t} \sin \omega_{d} t$
Solution: \[
Y(s)=\boxed{\frac{\omega_{d}}{(s+\sigma)^{2}+\omega_{d}^{2}}}
\]
Final answer: The final answer is \frac{\omega_{d}}{(s+\sigma)^{2}+\omega_{d}^{2}}. I hope it is correct.
Subproblem 2: $y(t)=e^{-\sigma t} \cos \omega_{d} t$ | \[
Y(s)=\boxed{\frac{s+\sigma}{(s+\sigma)^{2}+\omega_{d}^{2}}}
\] | Dynamics and Control (2.003 Spring 2005) | 117 |
Preamble: For each of the functions $y(t)$, find the Laplace Transform $Y(s)$ :
Subproblem 0: $y(t)=e^{-a t}$
Solution: This function is one of the most widely used in dynamic systems, so we memorize its transform!
\[
Y(s)=\boxed{\frac{1}{s+a}}
\]
Final answer: The final answer is \frac{1}{s+a}. I hope it is correct.
Subproblem 1: $y(t)=e^{-\sigma t} \sin \omega_{d} t$ | \[
Y(s)=\boxed{\frac{\omega_{d}}{(s+\sigma)^{2}+\omega_{d}^{2}}}
\] | Dynamics and Control (2.003 Spring 2005) | 118 |
Preamble: Consider the mass \(m\) sliding horizontally under the influence of the applied force \(f\) and a friction force which can be approximated by a linear friction element with coefficient \(b\).
Formulate the state-determined equation of motion for the velocity \(v\) as output and the force \(f\) as input. | The equation of motion is
\[
\boxed{m \frac{d v}{d t}+b v=f} \quad \text { or } \quad \frac{d v}{d t}=-\frac{b}{m} v+\frac{1}{m} f
\] | Dynamics and Control (2.003 Spring 2005) | 119 |
Preamble: Consider the rotor with moment of inertia \(I\) rotating under the influence of an applied torque \(T\) and the frictional torques from two bearings, each of which can be approximated by a linear frictional element with coefficient \(B\).
Subproblem 0: Formulate the state-determined equation of motion for the angular velocity $\omega$ as output and the torque $T$ as input.
Solution: The equation of motion is
\[
\boxed{I \frac{d \omega}{d t}+2 B \omega=T} \quad \text { or } \quad \frac{d \omega}{d t}=-\frac{2 B}{I} \omega+\frac{1}{I} T
\]
Final answer: The final answer is I \frac{d \omega}{d t}+2 B \omega=T. I hope it is correct.
Subproblem 1: Consider the case where:
\[
\begin{aligned}
I &=0.001 \mathrm{~kg}-\mathrm{m}^{2} \\
B &=0.005 \mathrm{~N}-\mathrm{m} / \mathrm{r} / \mathrm{s}
\end{aligned}
\]
What is the steady-state velocity \(\omega_{s s}\), in radians per second, when the input is a constant torque of 10 Newton-meters? | The steady-state angular velocity, when \(T=10\) Newton-meters, and \(I=0.001 \mathrm{~kg}-\mathrm{m}^{2}\), and \(B=0.005 \mathrm{~N}-\mathrm{m} / \mathrm{r} / \mathrm{s}\) is
\[
\omega_{s s}=\frac{T}{2 B}=\frac{10}{2(0.005)}=\boxed{1000} \mathrm{r} / \mathrm{s}
\] | Dynamics and Control (2.003 Spring 2005) | 120 |
Preamble: Consider the mass \(m\) sliding horizontally under the influence of the applied force \(f\) and a friction force which can be approximated by a linear friction element with coefficient \(b\).
Subproblem 0: Formulate the state-determined equation of motion for the velocity \(v\) as output and the force \(f\) as input.
Solution: The equation of motion is
\[
\boxed{m \frac{d v}{d t}+b v=f} \quad \text { or } \quad \frac{d v}{d t}=-\frac{b}{m} v+\frac{1}{m} f
\]
Final answer: The final answer is m \frac{d v}{d t}+b v=f. I hope it is correct.
Subproblem 1: Consider the case where:
\[
\begin{aligned}
m &=1000 \mathrm{~kg} \\
b &=100 \mathrm{~N} / \mathrm{m} / \mathrm{s}
\end{aligned}
\]
What is the steady-state velocity \(v_{s s}\) when the input is a constant force of 10 Newtons? Answer in meters per second. | The steady-state velocity, when \(f=10\) Newtons, and \(m=1000 \mathrm{~kg}\), and \(b=100 \mathrm{~N} / \mathrm{m} / \mathrm{s}\) is
\[
v_{s s}=\frac{f}{b}=\frac{10}{100}=\boxed{0.10} \mathrm{~m} / \mathrm{s}
\] | Dynamics and Control (2.003 Spring 2005) | 121 |
Obtain the inverse Laplace transform of the following frequency-domain expression: $F(s) = -\frac{(4 s-10)}{s(s+2)(s+5)}$.
Use $u(t)$ to denote the unit step function. | Using partial fraction expansion, the above can be rewritten as
\[
F(s) = \frac{1}{s} - \frac{3}{s+2} + \frac{2}{s+5}
\]
Apply the inverse Laplace transform, then we end up with
\[
f(t) = \boxed{(1 - 3e^{-2t} + 2e^{-5t}) u(t)}
\] | Dynamics and Control (2.003 Spring 2005) | 122 |
A signal has a Laplace transform
\[
X(s)=b+\frac{a}{s(s+a)}
\]
where \(a, b>0\), and with a region of convergence of \(|s|>0\). Find \(x(t), t>0\). | Each term of \(X(s)\) can be evaluated directly using a table of Laplace Transforms:
\[
\mathcal{L}^{-1}\{b\}=b \delta(t)
\]
and
\[
\mathcal{L}^{-1}\left\{\frac{a}{s(s+a)}\right\}=1-e^{-a t}
\]
The final result is then
\[
\mathcal{L}^{-1}\{X(s)\}=\boxed{b \delta(t)+1-e^{-a t}}
\] | Dynamics and Control (2.003 Spring 2005) | 123 |
Preamble: For each Laplace Transform \(Y(s)\), find the function \(y(t)\) :
\[
Y(s)=\boxed{\frac{1}{(s+a)(s+b)}}
\] | We can simplify with partial fractions:
\[
Y(s)=\frac{1}{(s+a)(s+b)}=\frac{C}{s+a}+\frac{D}{s+b}
\]
find the constants \(C\) and \(D\) by setting \(s=-a\) and \(s=-b\)
\[
\begin{aligned}
\frac{1}{(s+a)(s+b)} &=\frac{C}{s+a}+\frac{D}{s+b} \\
1 &=C(s+b)+D(s+a) \\
C &=\frac{1}{b-a} \\
D &=\frac{1}{a-b}
\end{aligned}
\]
therefore
\[
Y(s)=\frac{1}{b-a} \frac{1}{s+a}-\frac{1}{b-a} \frac{1}{s+b}
\]
By looking up the inverse Laplace Transform of \(\frac{1}{s+b}\), we find the total solution \(y(t)\)
\[
y(t)=\boxed{\frac{1}{b-a}\left(e^{-a t}-e^{-b t}\right)}
\] | Dynamics and Control (2.003 Spring 2005) | 124 |
Preamble: Consider the rotor with moment of inertia \(I\) rotating under the influence of an applied torque \(T\) and the frictional torques from two bearings, each of which can be approximated by a linear frictional element with coefficient \(B\).
Formulate the state-determined equation of motion for the angular velocity $\omega$ as output and the torque $T$ as input. | The equation of motion is
\[
\boxed{I \frac{d \omega}{d t}+2 B \omega=T} \quad \text { or } \quad \frac{d \omega}{d t}=-\frac{2 B}{I} \omega+\frac{1}{I} T
\] | Dynamics and Control (2.003 Spring 2005) | 125 |
Obtain the inverse Laplace transform of the following frequency-domain expression: $F(s) = \frac{4}{s^2(s^2+4)}$.
Use $u(t)$ to denote the unit step function. | Since $F(s) = \frac{1}{s^2} + \frac{-1}{s^2+4}$, its inverse Laplace transform is
\[
f(t) = \boxed{(t + \frac{1}{2} \sin{2t}) u(t)}
\] | Dynamics and Control (2.003 Spring 2005) | 126 |
Preamble: This problem considers the simple RLC circuit, in which a voltage source $v_{i}$ is in series with a resistor $R$, inductor $L$, and capacitor $C$. We measure the voltage $v_{o}$ across the capacitor. $v_{i}$ and $v_{o}$ share a ground reference.
Calculate the transfer function \(V_{o}(s) / V_{i}(s)\). | Using the voltage divider relationship:
\[
\begin{aligned}
V_{o}(s) &=\frac{Z_{e q}}{Z_{\text {total }}}V_{i}(s)=\frac{\frac{1}{C s}}{R+L s+\frac{1}{C s}} V_{i}(s) \\
\frac{V_{o}(s)}{V_{i}(s)} &=\boxed{\frac{1}{L C s^{2}+R C s+1}}
\end{aligned}
\] | Dynamics and Control (2.003 Spring 2005) | 127 |
Preamble: You are given an equation of motion of the form:
\[
\dot{y}+5 y=10 u
\]
What is the time constant for this system? | We find the homogenous solution, solving:
\[
\dot{y}+5 y=0
\]
by trying a solution of the form $y=A \cdot e^{s, t}$.
Calculation:
\[
\dot{y}=A \cdot s \cdot e^{s \cdot t} \mid \Rightarrow A \cdot s \cdot e^{s t}+5 A \cdot e^{s t}=0
\]
yields that $s=-5$, meaning the solution is $y=A \cdot e^{-5 \cdot t}=A \cdot e^{-t / \tau}$, meaning $\tau = \boxed{0.2}$. | Dynamics and Control (2.003 Spring 2005) | 128 |
Preamble: This problem considers the simple RLC circuit, in which a voltage source $v_{i}$ is in series with a resistor $R$, inductor $L$, and capacitor $C$. We measure the voltage $v_{o}$ across the capacitor. $v_{i}$ and $v_{o}$ share a ground reference.
Subproblem 0: Calculate the transfer function \(V_{o}(s) / V_{i}(s)\).
Solution: Using the voltage divider relationship:
\[
\begin{aligned}
V_{o}(s) &=\frac{Z_{e q}}{Z_{\text {total }}}V_{i}(s)=\frac{\frac{1}{C s}}{R+L s+\frac{1}{C s}} V_{i}(s) \\
\frac{V_{o}(s)}{V_{i}(s)} &=\boxed{\frac{1}{L C s^{2}+R C s+1}}
\end{aligned}
\]
Final answer: The final answer is \frac{1}{L C s^{2}+R C s+1}. I hope it is correct.
Subproblem 1: Let \(L=0.01 \mathrm{H}\). Choose the value of $C$ such that \(\omega_{n}=10^{5}\) and \(\zeta=0.05\). Give your answer in Farads. | $C=\frac{1}{\omega_{n}^{2}L}=\boxed{1e-8}[\mathrm{~F}]$ | Dynamics and Control (2.003 Spring 2005) | 129 |
Preamble: Here we consider a system described by the differential equation
\[
\ddot{y}+10 \dot{y}+10000 y=0 .
\]
What is the value of the natural frequency \(\omega_{n}\) in radians per second? | $\omega_{n}=\sqrt{\frac{k}{m}}$
So
$\omega_{n} =\boxed{100} \mathrm{rad} / \mathrm{s}$ | Dynamics and Control (2.003 Spring 2005) | 130 |
Preamble: Consider a circuit in which a voltage source of voltage in $v_{i}(t)$ is connected in series with an inductor $L$ and capacitor $C$. We consider the voltage across the capacitor $v_{o}(t)$ to be the output of the system.
Both $v_{i}(t)$ and $v_{o}(t)$ share ground reference.
Write the governing differential equation for this circuit. | Using Kirchoff Current Law at the node between the inductor and capacitor with the assumed currents both positive into the node gives the following:
\[
\begin{gathered}
i_{L}+i_{C}=0 \\
i_{L}=\frac{1}{L} \int v_{L} d t \\
i_{C}=C \frac{d v_{c}}{d t}
\end{gathered}
\]
The above equation must be differentiated before substituting for the currents and from the direction of our assumed currents, \(v_{L}=v_{i}-v_{o}\) and \(v_{C}=0-v_{o}\). The governing differential equation is then
\[
\boxed{\frac{d^{2} v_{o}}{d t^{2}}+\frac{v_{o}}{L C}=\frac{v_{i}}{L C}}
\] | Dynamics and Control (2.003 Spring 2005) | 131 |
Write (but don't solve) the equation of motion for a pendulum consisting of a mass $m$ attached to a rigid massless rod, suspended from the ceiling and free to rotate in a single vertical plane. Let the rod (of length $l$) make an angle of $\theta$ with the vertical. Gravity ($mg$) acts directly downward, the system input is a horizontal external force $f(t)$, and the system output is the angle $\theta(t)$.
Note: Do NOT make the small-angle approximation in your equation. | From force balance, we can derive the equation of motion. Choosing the system variable system variable $\theta(t)$ with polar coordinates, we don't need to care about tension on the rod and centrifugal force.
We can use the relation between torque and angular momentum to immediately write down the equation for $\theta(t)$:
\[
m l^{2} \ddot{\theta}(t)-m g l \sin \theta(t)=f(t) l \cos \theta(t) .
\]
Dividing both sides by $l$ gives:
\[
\boxed{m l \ddot{\theta}(t)-m g \sin \theta(t)=f(t) \cos \theta(t)} .
\]
Note that inertia of the mass with respect to the rotation axis is $m l^{2}$. It is a non linear differential equation because it has $\sin \theta(t)$ term. | Dynamics and Control (2.003 Spring 2005) | 132 |
Preamble: Here we consider a system described by the differential equation
\[
\ddot{y}+10 \dot{y}+10000 y=0 .
\]
Subproblem 0: What is the value of the natural frequency \(\omega_{n}\) in radians per second?
Solution: $\omega_{n}=\sqrt{\frac{k}{m}}$
So
$\omega_{n} =\boxed{100} \mathrm{rad} / \mathrm{s}$
Final answer: The final answer is 100. I hope it is correct.
Subproblem 1: What is the value of the damping ratio \(\zeta\)?
Solution: $\zeta=\frac{b}{2 \sqrt{k m}}$
So
$\zeta =\boxed{0.05}$
Final answer: The final answer is 0.05. I hope it is correct.
Subproblem 2: What is the value of the damped natural frequency \(\omega_{d}\) in radians per second? Give your answer to three significant figures. | $\omega_{d}=\omega_{n} \sqrt{1-\zeta^{2}}$
So
$\omega_{d}=\boxed{99.9} \mathrm{rad} / \mathrm{s}$ | Dynamics and Control (2.003 Spring 2005) | 133 |
Preamble: Here we consider a system described by the differential equation
\[
\ddot{y}+10 \dot{y}+10000 y=0 .
\]
Subproblem 0: What is the value of the natural frequency \(\omega_{n}\) in radians per second?
Solution: $\omega_{n}=\sqrt{\frac{k}{m}}$
So
$\omega_{n} =\boxed{100} \mathrm{rad} / \mathrm{s}$
Final answer: The final answer is 100. I hope it is correct.
Subproblem 1: What is the value of the damping ratio \(\zeta\)? | $\zeta=\frac{b}{2 \sqrt{k m}}$
So
$\zeta =\boxed{0.05}$ | Dynamics and Control (2.003 Spring 2005) | 134 |
What is the speed of light in meters/second to 1 significant figure? Use the format $a \times 10^{b}$ where a and b are numbers. | $\boxed{3e8}$ m/s. | Relativity (8.033 Fall 2006) | 135 |
Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested.
Subproblem 0: Age of our universe when most He nuclei were formed in minutes:
Solution: \boxed{1} minute.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: Age of our universe when hydrogen atoms formed in years:
Solution: \boxed{400000} years.
Final answer: The final answer is 400000. I hope it is correct.
Subproblem 2: Age of our universe today in Gyr:
Solution: \boxed{10} Gyr.
Final answer: The final answer is 10. I hope it is correct.
Subproblem 3: Number of stars in our Galaxy: (Please format your answer as 'xen' representing $x * 10^n$) | \boxed{1e11}. | Relativity (8.033 Fall 2006) | 136 |
Preamble: In a parallel universe, the Boston baseball team made the playoffs.
Manny Relativirez hits the ball and starts running towards first base at speed $\beta$. How fast is he running, given that he sees third base $45^{\circ}$ to his left (as opposed to straight to his left before he started running)? Assume that he is still very close to home plate. Give your answer in terms of the speed of light, $c$. | Using the aberration formula with $\cos \theta^{\prime}=-1 / \sqrt{2}, \beta=1 / \sqrt{2}$, so $v=\boxed{\frac{1}{\sqrt{2}}c}$. | Relativity (8.033 Fall 2006) | 137 |
Preamble: In the Sun, one of the processes in the He fusion chain is $p+p+e^{-} \rightarrow d+\nu$, where $d$ is a deuteron. Make the approximations that the deuteron rest mass is $2 m_{p}$, and that $m_{e} \approx 0$ and $m_{\nu} \approx 0$, since both the electron and the neutrino have negligible rest mass compared with the proton rest mass $m_{p}$.
In the lab frame, the two protons have the same energy $\gamma m_{p}$ and impact angle $\theta$, and the electron is at rest. Calculate the energy $E_{\nu}$ of the neutrino in the rest frame of the deuteron in terms of $\theta, m_{p}$ and $\gamma$. | Use the fact that the quantity $E^{2}-p^{2} c^{2}$ is invariant. In the deutron's rest frame, after the collison:
\[
\begin{aligned}
E^{2}-p^{2} c^{2} &=\left(2 m_{p} c^{2}+E_{\nu}\right)^{2}-E_{\nu}^{2} \\
&=4 m_{p}^{2} c^{4}+4 m_{p} c^{2} E_{\nu}=4 m_{p} c^{2}\left(m_{p} c^{2}+E_{\nu}\right)
\end{aligned}
\]
In the lab frame, before collison:
\[
\begin{aligned}
E^{2}-p^{2} c^{2} &=\left(2 E_{p}\right)^{2}-\left(2 p_{p} \cos \theta c\right)^{2} \\
&=\left(2 \gamma m_{p} c^{2}\right)^{2}-\left(2 \gamma \beta m_{p} \cos \theta c^{2}\right)^{2}
\end{aligned}
\]
Use $\gamma^{2} \beta^{2}=\left(\gamma^{2}-1\right)$ in the second term and simplify the algebra to find
\[
E^{2}-p^{2} c^{2}=4 m_{p}^{2} c^{4}\left(\gamma^{2}-\left(\gamma^{2}-1\right) \cos ^{2} \theta\right)
\]
Equating the invariants in the two frames, we have
\[
\begin{aligned}
4 m_{p} c^{2}\left(m_{p} c^{2}+E_{\nu}\right) &=4 m_{p}^{2} c^{4}\left(\gamma^{2}-\left(\gamma^{2}-1\right) \cos ^{2} \theta\right) \\
\Rightarrow E_{\nu} &= \boxed{m_{p} c^{2}\left(\gamma^{2}-1\right) \sin ^{2} \theta}
\end{aligned}
\] | Relativity (8.033 Fall 2006) | 138 |
Preamble: In a parallel universe, the Boston baseball team made the playoffs.
Subproblem 0: Manny Relativirez hits the ball and starts running towards first base at speed $\beta$. How fast is he running, given that he sees third base $45^{\circ}$ to his left (as opposed to straight to his left before he started running)? Assume that he is still very close to home plate. Give your answer in terms of the speed of light, $c$.
Solution: Using the aberration formula with $\cos \theta^{\prime}=-1 / \sqrt{2}, \beta=1 / \sqrt{2}$, so $v=\boxed{\frac{1}{\sqrt{2}}c}$.
Final answer: The final answer is \frac{1}{\sqrt{2}}c. I hope it is correct.
Subproblem 1: A player standing on third base is wearing red socks emitting light of wavelength $\lambda_{\text {red}}$. What wavelength does Manny see in terms of $\lambda_{\text {red}}$? | Using the doppler shift formula, $\lambda^{\prime}= \boxed{\lambda_{\text {red}} / \sqrt{2}}$. | Relativity (8.033 Fall 2006) | 139 |
Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested.
Subproblem 0: Age of our universe when most He nuclei were formed in minutes:
Solution: \boxed{1} minute.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: Age of our universe when hydrogen atoms formed in years:
Solution: \boxed{400000} years.
Final answer: The final answer is 400000. I hope it is correct.
Subproblem 2: Age of our universe today in Gyr: | \boxed{10} Gyr. | Relativity (8.033 Fall 2006) | 140 |
How many down quarks does a tritium ($H^3$) nucleus contain? | \boxed{5}. | Relativity (8.033 Fall 2006) | 141 |
How many up quarks does a tritium ($H^3$) nucleus contain? | \boxed{4}. | Relativity (8.033 Fall 2006) | 142 |
Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested.
Age of our universe when most He nuclei were formed in minutes: | \boxed{1} minute. | Relativity (8.033 Fall 2006) | 143 |
Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested.
Subproblem 0: Age of our universe when most He nuclei were formed in minutes:
Solution: \boxed{1} minute.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: Age of our universe when hydrogen atoms formed in years:
Solution: \boxed{400000} years.
Final answer: The final answer is 400000. I hope it is correct.
Subproblem 2: Age of our universe today in Gyr:
Solution: \boxed{10} Gyr.
Final answer: The final answer is 10. I hope it is correct.
Subproblem 3: Number of stars in our Galaxy: (Please format your answer as 'xen' representing $x * 10^n$)
Solution: \boxed{1e11}.
Final answer: The final answer is 1e11. I hope it is correct.
Subproblem 4: Light travel time to closest star (Sun!:) in minutes. (Please format your answer as an integer.) | \boxed{8} minutes. | Relativity (8.033 Fall 2006) | 144 |
Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested.
Subproblem 0: Age of our universe when most He nuclei were formed in minutes:
Solution: \boxed{1} minute.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: Age of our universe when hydrogen atoms formed in years: | \boxed{400000} years. | Relativity (8.033 Fall 2006) | 145 |
Potassium metal can be used as the active surface in a photodiode because electrons are relatively easily removed from a potassium surface. The energy needed is $2.15 \times 10^{5} J$ per mole of electrons removed ( 1 mole $=6.02 \times 10^{23}$ electrons). What is the longest wavelength light (in nm) with quanta of sufficient energy to eject electrons from a potassium photodiode surface? | \includegraphics[scale=0.5]{set_02_img_00.jpg}
\nonessentialimage
$I_{p}$, the photocurrent, is proportional to the intensity of incident radiation, i.e. the number of incident photons capable of generating a photoelectron.
This device should be called a phototube rather than a photodiode - a solar cell is a photodiode.
Required: $1 eV=1.6 \times 10^{-19} J$
\[
E_{\text {rad }}=h v=(hc) / \lambda
\]
The question is: below what threshold energy (hv) will a photon no longer be able to generate a photoelectron?\\
$2.15 x 10^{5}$ J/mole photoelectrons $\times \frac{1 \text{mole}}{6.02 \times 10^{23} \text{photoelectrons}} = 3.57 \times 10^{-19}$ J/photoelectron\\
$\lambda_{\text {threshold }}=\frac{hc}{3.57 \times 10^{-19}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{3.57 \times 10^{-19}}=5.6 \times 10^{-7} m= \boxed{560} nm$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 146 |
Preamble: For red light of wavelength $(\lambda) 6.7102 \times 10^{-5} cm$, emitted by excited lithium atoms, calculate:
Subproblem 0: the frequency $(v)$ in Hz, to 4 decimal places.
Solution: $c=\lambda v$ and $v=c / \lambda$ where $v$ is the frequency of radiation (number of waves/s).
For: $\quad \lambda=6.7102 \times 10^{-5} cm=6.7102 \times 10^{-7} m$
\[
v=\frac{2.9979 \times 10^{8} {ms}^{-1}}{6.7102 \times 10^{-7} m}=4.4677 \times 10^{14} {s}^{-1}= \boxed{4.4677} Hz
\]
Final answer: The final answer is 4.4677. I hope it is correct.
Subproblem 1: the wave number $(\bar{v})$ in ${cm}^{-1}$. Please format your answer as $n \times 10^x$, where $n$ is to 4 decimal places.
Solution: $\bar{v}=\frac{1}{\lambda}=\frac{1}{6.7102 \times 10^{-7} m}=1.4903 \times 10^{6} m^{-1}= \boxed{1.4903e4} {cm}^{-1}$
Final answer: The final answer is 1.4903e4. I hope it is correct.
Subproblem 2: the wavelength $(\lambda)$ in nm, to 2 decimal places. | $\lambda=6.7102 \times 10^{-5} cm \times \frac{1 nm}{10^{-7} cm}= \boxed{671.02} cm$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 147 |
What is the net charge of arginine in a solution of $\mathrm{pH} \mathrm{} 1.0$ ? Please format your answer as +n or -n. | \boxed{+2}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 148 |
Preamble: For red light of wavelength $(\lambda) 6.7102 \times 10^{-5} cm$, emitted by excited lithium atoms, calculate:
Subproblem 0: the frequency $(v)$ in Hz, to 4 decimal places.
Solution: $c=\lambda v$ and $v=c / \lambda$ where $v$ is the frequency of radiation (number of waves/s).
For: $\quad \lambda=6.7102 \times 10^{-5} cm=6.7102 \times 10^{-7} m$
\[
v=\frac{2.9979 \times 10^{8} {ms}^{-1}}{6.7102 \times 10^{-7} m}=4.4677 \times 10^{14} {s}^{-1}= \boxed{4.4677} Hz
\]
Final answer: The final answer is 4.4677. I hope it is correct.
Subproblem 1: the wave number $(\bar{v})$ in ${cm}^{-1}$. Please format your answer as $n \times 10^x$, where $n$ is to 4 decimal places. | $\bar{v}=\frac{1}{\lambda}=\frac{1}{6.7102 \times 10^{-7} m}=1.4903 \times 10^{6} m^{-1}= \boxed{1.4903e4} {cm}^{-1}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 149 |
Determine the atomic weight of ${He}^{++}$ in amu to 5 decimal places from the values of its constituents. | The mass of the constituents $(2 p+2 n)$ is given as:
\[
\begin{array}{ll}
2 p= & 2 \times 1.6726485 \times 10^{-24} g \\
2 n= & 2 \times 16749543 \times 10^{-24} g
\end{array}
\]
The atomic weight (calculated) in amu is given as:
\[
\begin{aligned}
&\frac{6.6952056 \times 10^{-24} g}{1.660565 \times 10^{-24} g} / amu \\
&{He}=\boxed{4.03188} amu
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 150 |
Preamble: Determine the following values from a standard radio dial.
Subproblem 0: What is the minimum wavelength in m for broadcasts on the AM band? Format your answer as an integer.
Solution: \[
\mathrm{c}=v \lambda, \therefore \lambda_{\min }=\frac{\mathrm{c}}{v_{\max }} ; \lambda_{\max }=\frac{\mathrm{c}}{v_{\min }}
\]
$\lambda_{\min }=\frac{3 \times 10^{8} m / s}{1600 \times 10^{3} Hz}=\boxed{188} m$
Final answer: The final answer is 188. I hope it is correct.
Subproblem 1: What is the maximum wavelength in m for broadcasts on the AM band? Format your answer as an integer. | \[
\mathrm{c}=v \lambda, \therefore \lambda_{\min }=\frac{\mathrm{c}}{v_{\max }} ; \lambda_{\max }=\frac{\mathrm{c}}{v_{\min }}
\]
\[
\lambda_{\max }=\frac{3 \times 10^{8}}{530 \times 10^{3}}=\boxed{566} m
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 151 |
Determine the wavelength of radiation emitted by hydrogen atoms in angstroms upon electron transitions from $n=6$ to $n=2$. | From the Rydberg relationship we obtain:
\[
\begin{aligned}
&\frac{1}{\lambda}=\bar{v}=R\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right)=1.097 \times 10^{7}\left(\frac{1}{36}-\frac{1}{4}\right)=(-) 2.44 \times 10^{6} \\
&\lambda=\frac{1}{v}=\frac{1}{2.44 \times 10^{6}}=4.1 \times 10^{-7} {~m}=0.41 \mu {m}=\boxed{4100} \text{angstroms}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 152 |
Preamble: Determine the following values from a standard radio dial.
Subproblem 0: What is the minimum wavelength in m for broadcasts on the AM band? Format your answer as an integer.
Solution: \[
\mathrm{c}=v \lambda, \therefore \lambda_{\min }=\frac{\mathrm{c}}{v_{\max }} ; \lambda_{\max }=\frac{\mathrm{c}}{v_{\min }}
\]
$\lambda_{\min }=\frac{3 \times 10^{8} m / s}{1600 \times 10^{3} Hz}=\boxed{188} m$
Final answer: The final answer is 188. I hope it is correct.
Subproblem 1: What is the maximum wavelength in m for broadcasts on the AM band? Format your answer as an integer.
Solution: \[
\mathrm{c}=v \lambda, \therefore \lambda_{\min }=\frac{\mathrm{c}}{v_{\max }} ; \lambda_{\max }=\frac{\mathrm{c}}{v_{\min }}
\]
\[
\lambda_{\max }=\frac{3 \times 10^{8}}{530 \times 10^{3}}=\boxed{566} m
\]
Final answer: The final answer is 566. I hope it is correct.
Subproblem 2: What is the minimum wavelength in m (to 2 decimal places) for broadcasts on the FM band? | \[
\mathrm{c}=v \lambda, \therefore \lambda_{\min }=\frac{\mathrm{c}}{v_{\max }} ; \lambda_{\max }=\frac{\mathrm{c}}{v_{\min }}
\]
$\lambda_{\min }=\frac{3 \times 10^{8}}{108 \times 10^{6}}=\boxed{2.78} m$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 153 |
Calculate the "Bohr radius" in angstroms to 3 decimal places for ${He}^{+}$. | In its most general form, the Bohr theory considers the attractive force (Coulombic) between the nucleus and an electron being given by:
\[
F_{c}=\frac{Z e^{2}}{4 \pi \varepsilon_{0} r^{2}}
\]
where Z is the charge of the nucleus ( 1 for H, 2 for He, etc.). Correspondingly, the electron energy $\left(E_{e l}\right)$ is given as:
\[
E_{e l}=-\frac{z^{2}}{n^{2}} \frac{m e^{4}}{8 h^{2} \varepsilon_{0}^{2}}
\]
and the electronic orbit $\left(r_{n}\right)$ :
\[
\begin{aligned}
&r_{n}=\frac{n^{2}}{Z} \frac{n^{2} \varepsilon_{0}}{\pi m e^{2}} \\
&r_{n}=\frac{n^{2}}{Z} a_{0}
\end{aligned}
\]
For ${He}^{+}(Z=2), {r}_{1}=\frac{1}{2} {a}_{0}=\frac{0.529}{2} \times 10^{-10} m=\boxed{0.264}$ angstroms | Introduction to Solid State Chemistry (3.091 Fall 2010) | 154 |
Preamble: For red light of wavelength $(\lambda) 6.7102 \times 10^{-5} cm$, emitted by excited lithium atoms, calculate:
the frequency $(v)$ in Hz, to 4 decimal places. | $c=\lambda v$ and $v=c / \lambda$ where $v$ is the frequency of radiation (number of waves/s).
For: $\quad \lambda=6.7102 \times 10^{-5} cm=6.7102 \times 10^{-7} m$
\[
v=\frac{2.9979 \times 10^{8} {ms}^{-1}}{6.7102 \times 10^{-7} m}=4.4677 \times 10^{14} {s}^{-1}= \boxed{4.4677} Hz
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 155 |
Electromagnetic radiation of frequency $3.091 \times 10^{14} \mathrm{~Hz}$ illuminates a crystal of germanium (Ge). Calculate the wavelength of photoemission in meters generated by this interaction. Germanium is an elemental semiconductor with a band gap, $E_{g}$, of $0.7 \mathrm{eV}$. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | First compare $E$ of the incident photon with $E_{g}$ :
\[
\begin{aligned}
&\mathrm{E}_{\text {incident }}=\mathrm{hv}=6.6 \times 10^{-34} \times 3.091 \times 10^{14}=2.04 \times 10^{-19} \mathrm{~J} \\
&\mathrm{E}_{\mathrm{g}}=0.7 \mathrm{eV}=1.12 \times 10^{-19} \mathrm{~J}<\mathrm{E}_{\text {incident }}
\end{aligned}
\]
$\therefore$ electron promotion is followed by emission of a new photon of energy equal to $E_{g}$, and energy in excess of $E_{g}$ is dissipated as heat in the crystal
\includegraphics[scale=0.5]{set_17_img_00.jpg}
\nonessentialimage
$$
\lambda_{\text {emitted }}=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{0.7 \times 1.6 \times 10^{-19}}= \boxed{1.77e-6} \mathrm{~m}
$$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 156 |
What is the energy gap (in eV, to 1 decimal place) between the electronic states $n=3$ and $n=8$ in a hydrogen atom? | \[
\begin{array}{rlr}
\text { Required: } & \Delta {E}_{{el}}=\left(\frac{1}{{n}_{{i}}^{2}}-\frac{1}{{n}_{{f}}^{2}}\right) {K} ; & {K}=2.18 \times 10^{-18} \\
& \text { Or } \bar{v}=\left(\frac{1}{{n}_{{i}}^{2}}-\frac{1}{{n}_{{f}}^{2}}\right) {R} ; & {R}=1.097 \times 10^{7} {~m}^{-1}
\end{array}
\]
(Since only the energy gap is asked, we are not concerned about the sign.)
\[
\begin{aligned}
&\Delta {E}=(1 / 9-1 / 65) {K}=0.0955 \times 2.18 \times 10^{-18} {~J} \\
&\Delta {E}=2.08 \times 10^{-19} {~J}=\boxed{1.3} {eV}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 157 |
Determine for hydrogen the velocity in m/s of an electron in an ${n}=4$ state. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | This problem may be solved in a variety of ways, the simplest of which makes use of the Bohr quantization of the angular momentum:
\[
\begin{aligned}
&m v r=n \times \frac{h}{2 \pi} \quad\left(r=r_{0} n^{2}\right) \\
&m v r_{0} n^{2}=n \times \frac{h}{2 \pi} \\
&v=\frac{h}{2 \pi m r_{0} n}= \boxed{5.47e5} m/s
\end{aligned}
\]
(A numerically correct result is obtained by taking:
\[
E_{e l}=-\frac{1}{n^{2}} K=\frac{m v^{2}}{2}
\]
The negative sign reflects the $E_{\text {pot }}$ term, which happens to be $-2 E_{K i n}$.) | Introduction to Solid State Chemistry (3.091 Fall 2010) | 158 |
Preamble: A pure crystalline material (no impurities or dopants are present) appears red in transmitted light.
Subproblem 0: Is this material a conductor, semiconductor or insulator? Give the reasons for your answer.
Solution: If the material is pure (no impurity states present), then it must be classified as a \boxed{semiconductor} since it exhibits a finite "band gap" - i.e. to activate charge carriers, photons with energies in excess of "red" radiation are required.
Final answer: The final answer is semiconductor. I hope it is correct.
Subproblem 1: What is the approximate band gap $\left(\mathrm{E}_{g}\right)$ for this material in eV? Please round your answer to 1 decimal place. | "White light" contains radiation in wavelength ranging from about $4000 \AA$ (violet) to $7000 \AA$ (deep red). A material appearing red in transmission has the following absorption characteristics:
\includegraphics[scale=0.5]{set_17_img_06.jpg}
\nonessentialimage
Taking $\lambda=6500 \AA$ as the optical absorption edge for this material, we have:
\[
E=\frac{\mathrm{hc}}{\lambda}=3.05 \times 10^{-29} \mathrm{~J} \times \frac{1 \mathrm{eV}}{1.6 \times 10^{-19} \mathrm{~J}}=1.9 \mathrm{eV}
\]
Accordingly, the band gap for the material is $E_{g}= \boxed{1.9} \mathrm{eV}$. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 159 |
Calculate the minimum potential $(V)$ in volts (to 1 decimal place) which must be applied to a free electron so that it has enough energy to excite, upon impact, the electron in a hydrogen atom from its ground state to a state of $n=5$. | We can picture this problem more clearly: an electron is accelerated by a potential, $V x$, and thus acquires the kinetic energy e $x V_{x}\left[=\left(m v^{2}\right) / 2\right.$ which is to be exactly the energy required to excite an electron in hydrogen from $n=1$ to $n=5$.\\
${e} \cdot {V}_{{x}} =-{K}\left(\frac{1}{25}-\frac{1}{1}\right) $\\
${V}_{{x}} =\frac{{K}}{{e}} \times \frac{24}{25}=\frac{2.18 \times 10^{-18}}{1.6 \times 10^{-19}} \times \frac{24}{25}= \boxed{13.1} {Volt}$ \\
${\left[13.1 {eV}=13.1 {eV} \times \frac{1.6 \times 10^{-19} {~J}}{{eV}}=2.08 \times 10^{-18} {~J}=-{K}\left(\frac{1}{{n}_{{f}}^{2}}-\frac{1}{{n}_{{i}}^{2}}\right)\right]}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 160 |
Preamble: For light with a wavelength $(\lambda)$ of $408 \mathrm{~nm}$ determine:
Subproblem 0: the frequency in $s^{-1}$. Please format your answer as $n \times 10^x$, where $n$ is to 3 decimal places.
Solution: To solve this problem we must know the following relationships:
\[
\begin{aligned}
v \lambda &=c
\end{aligned}
\]
$v$ (frequency) $=\frac{c}{\lambda}=\frac{3 \times 10^{8} m / s}{408 \times 10^{-9} m}= \boxed{7.353e14} s^{-1}$
Final answer: The final answer is 7.353e14. I hope it is correct.
Subproblem 1: the wave number in $m^{-1}$. Please format your answer as $n \times 10^x$, where $n$ is to 2 decimal places.
Solution: To solve this problem we must know the following relationships:
\[
\begin{aligned}
1 / \lambda=\bar{v}
\end{aligned}
\]
$\bar{v}$ (wavenumber) $=\frac{1}{\lambda}=\frac{1}{408 \times 10^{-9} m}=\boxed{2.45e6} m^{-1}$
Final answer: The final answer is 2.45e6. I hope it is correct.
Subproblem 2: the wavelength in angstroms. | To solve this problem we must know the following relationships:
\[
\begin{aligned}
m =10^{10} angstrom
\end{aligned}
\]
$\lambda=408 \times 10^{-9} m \times \frac{10^{10} angstrom}{\mathrm{m}}=\boxed{4080} angstrom$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 161 |
Preamble: Reference the information below to solve the following problems.
$\begin{array}{llll}\text { Element } & \text { Ionization Potential } & \text { Element } & \text { Ionization Potential } \\ {Na} & 5.14 & {Ca} & 6.11 \\ {Mg} & 7.64 & {Sc} & 6.54 \\ {Al} & 5.98 & {Ti} & 6.82 \\ {Si} & 8.15 & {~V} & 6.74 \\ {P} & 10.48 & {Cr} & 6.76 \\ {~S} & 10.36 & {Mn} & 7.43 \\ {Cl} & 13.01 & {Fe} & 7.9 \\ {Ar} & 15.75 & {Co} & 7.86 \\ & & {Ni} & 7.63 \\ & & {Cu} & 7.72\end{array}$
Subproblem 0: What is the first ionization energy (in J, to 3 decimal places) for Na?
Solution: The required data can be obtained by multiplying the ionization potentials (listed in the Periodic Table) with the electronic charge ( ${e}^{-}=1.6 \times 10^{-19}$ C).
\boxed{0.822} J.
Final answer: The final answer is 0.822. I hope it is correct.
Subproblem 1: What is the first ionization energy (in J, to 2 decimal places) for Mg? | The required data can be obtained by multiplying the ionization potentials (listed in the Periodic Table) with the electronic charge ( ${e}^{-}=1.6 \times 10^{-19}$ C).
\boxed{1.22} J. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 162 |
Light of wavelength $\lambda=4.28 \times 10^{-7} {~m}$ interacts with a "motionless" hydrogen atom. During this interaction it transfers all its energy to the orbiting electron of the hydrogen. What is the velocity in m/s of this electron after interaction? Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | First of all, a sketch:
\includegraphics[scale=0.5]{set_03_img_00.jpg}
\nonessentialimage
\[
\begin{aligned}
&\text { possibly to } {n}=\infty \text { (ionization), } \\
&\text { depending on the magnitude of } E(h v)
\end{aligned}
\]
let us see: $E(h v)=(h c) / \lambda=4.6 \times 10^{-19} {~J}$
To move the electron from $n=1$ to $n=2$ (minimum energy required for absorption of the photon), we have:
\[
\begin{aligned}
\Delta {E}=\left(\frac{1}{{n}_{{i}}^{2}}-\frac{1}{{n}_{{f}}^{2}}\right) {K} &=\frac{3}{4} {~K} \\
&=\frac{3}{4} \times 2.18 \times 10^{-18} {~J}=1.6 \times 10^{-18} {~J}
\end{aligned}
\]
We recognize that the photon energy is less than the $\Delta E_{\min }$ (for $n=1 \rightarrow n=2$ ).
This means that no interaction can take place - the photon will "pass by" and the electron will continue to orbit in its $1 s$ state! Its orbiting velocity can be obtained from:
\[
\begin{aligned}
&m v r=n\left(\frac{h}{2 \pi}\right) \\
&v=n\left(\frac{h}{2 \pi m r}\right)= \boxed{2.19e6} {~m} / {s}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 163 |
Determine the minimum potential in V (to 2 decimal places) that must be applied to an $\alpha$-particle so that on interaction with a hydrogen atom, a ground state electron will be excited to $n$ $=6$. | \[
\Delta {E}_{1 \rightarrow 6}={qV} \quad \therefore {V}=\frac{\Delta {E}_{1 \rightarrow 6}}{{q}}
\]
\[
\begin{aligned}
& \Delta {E}_{1 \rightarrow 6}=-{K}\left(\frac{1}{1^{2}}-\frac{1}{6^{2}}\right)=\frac{35}{36} {K} \\
& {q}=+2 {e} \\
& \therefore \quad V=\frac{35}{36} \times \frac{2.18 \times 10^{18}}{2 \times 1.6 \times 10^{-19}}=\boxed{6.62} V
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 164 |
Preamble: Reference the information below to solve the following problems.
$\begin{array}{llll}\text { Element } & \text { Ionization Potential } & \text { Element } & \text { Ionization Potential } \\ {Na} & 5.14 & {Ca} & 6.11 \\ {Mg} & 7.64 & {Sc} & 6.54 \\ {Al} & 5.98 & {Ti} & 6.82 \\ {Si} & 8.15 & {~V} & 6.74 \\ {P} & 10.48 & {Cr} & 6.76 \\ {~S} & 10.36 & {Mn} & 7.43 \\ {Cl} & 13.01 & {Fe} & 7.9 \\ {Ar} & 15.75 & {Co} & 7.86 \\ & & {Ni} & 7.63 \\ & & {Cu} & 7.72\end{array}$
What is the first ionization energy (in J, to 3 decimal places) for Na? | The required data can be obtained by multiplying the ionization potentials (listed in the Periodic Table) with the electronic charge ( ${e}^{-}=1.6 \times 10^{-19}$ C).
\boxed{0.822} J. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 165 |
Preamble: For "yellow radiation" (frequency, $v,=5.09 \times 10^{14} s^{-1}$ ) emitted by activated sodium, determine:
Subproblem 0: the wavelength $(\lambda)$ in m. Please format your answer as $n \times 10^x$, where n is to 2 decimal places.
Solution: The equation relating $v$ and $\lambda$ is $c=v \lambda$ where $c$ is the speed of light $=3.00 \times 10^{8} \mathrm{~m}$.
\[
\lambda=\frac{c}{v}=\frac{3.00 \times 10^{8} m / s}{5.09 \times 10^{14} s^{-1}}=\boxed{5.89e-7} m
\]
Final answer: The final answer is 5.89e-7. I hope it is correct.
Subproblem 1: the wave number $(\bar{v})$ in ${cm}^{-1}$. Please format your answer as $n \times 10^x$, where n is to 2 decimal places. | The wave number is $1 /$ wavelength, but since the wavelength is in m, and the wave number should be in ${cm}^{-1}$, we first change the wavelength into cm :
\[
\lambda=5.89 \times 10^{-7} m \times 100 cm / m=5.89 \times 10^{-5} cm
\]
Now we take the reciprocal of the wavelength to obtain the wave number:
\[
\bar{v}=\frac{1}{\lambda}=\frac{1}{5.89 \times 10^{-5} cm}= \boxed{1.70e4} {cm}^{-1}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 166 |
Subproblem 0: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{CO}$?
Solution: \boxed{1}.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{O}_{2}$ (in decimal form)? | \boxed{0.5}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 167 |
Preamble: Calculate the molecular weight in g/mole (to 2 decimal places) of each of the substances listed below.
$\mathrm{NH}_{4} \mathrm{OH}$ | $\mathrm{NH}_{4} \mathrm{OH}$ :
$5 \times 1.01=5.05(\mathrm{H})$
$1 \times 14.01=14.01(\mathrm{~N})$
$1 \times 16.00=16.00(\mathrm{O})$
$\mathrm{NH}_{4} \mathrm{OH}= \boxed{35.06}$ g/mole | Introduction to Solid State Chemistry (3.091 Fall 2010) | 168 |
Subproblem 0: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{CO}$?
Solution: \boxed{1}.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{O}_{2}$ (in decimal form)?
Solution: \boxed{0.5}.
Final answer: The final answer is 0.5. I hope it is correct.
Subproblem 2: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{CO}_{2}$ (in decimal form)? | \boxed{1}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 169 |
Magnesium (Mg) has the following isotopic distribution:
\[
\begin{array}{ll}
24_{\mathrm{Mg}} & 23.985 \mathrm{amu} \text { at } 0.7870 \text { fractional abundance } \\
25_{\mathrm{Mg}} & 24.986 \mathrm{amu} \text { at } 0.1013 \text { fractional abundance } \\
26_{\mathrm{Mg}} & 25.983 \mathrm{amu} \text { at } 0.1117 \text { fractional abundance }
\end{array}
\]
What is the atomic weight of magnesium (Mg) (to 3 decimal places) according to these data? | The atomic weight is the arithmetic average of the atomic weights of the isotopes, taking into account the fractional abundance of each isotope.
\[
\text { At.Wt. }=\frac{23.985 \times 0.7870+24.986 \times 0.1013+25.983 \times 0.1117}{0.7870+0.1013+0.1117}=\boxed{24.310}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 170 |
Preamble: Electrons are accelerated by a potential of 10 Volts.
Determine their velocity in m/s. Please format your answer as $n \times 10^x$, where $n$ is to 2 decimal places. | The definition of an ${eV}$ is the energy gained by an electron when it is accelerated through a potential of $1 {~V}$, so an electron accelerated by a potential of $10 {~V}$ would have an energy of $10 {eV}$.\\
${E}=\frac{1}{2} m {v}^{2} \rightarrow {v}=\sqrt{2 {E} / {m}}$
\[
E=10 {eV}=1.60 \times 10^{-18} {~J}
\]
\[
\begin{aligned}
& {m}=\text { mass of electron }=9.11 \times 10^{-31} {~kg} \\
& v=\sqrt{\frac{2 \times 1.6 \times 10^{-18} {~J}}{9.11 \times 10^{-31} {~kg}}}= \boxed{1.87e6} {~m} / {s}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 171 |
Determine the frequency (in $s^{-1}$ of radiation capable of generating, in atomic hydrogen, free electrons which have a velocity of $1.3 \times 10^{6} {~ms}^{-1}$. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | Remember the ground state electron energy in hydrogen $\left({K}=-2.18 \times 10^{-18} {~J}\right)$. The radiation in question will impart to the removed electron a velocity of $1.3 {x}$ $10^{6} {~ms}^{-1}$, which corresponds to:
\[
\begin{aligned}
&E_{\text {Kin }}=\frac{m v^{2}}{2}=\frac{9.1 \times 10^{-31} \times\left(1.3 \times 10^{6}\right)^{2}}{2} \text { Joules }=7.69 \times 10^{-19} {~J} \\
&E_{\text {rad }}=E_{\text {Kin }}+E_{\text {ioniz }}=7.69 \times 10^{-19}+2.18 \times 10^{-18}=2.95 \times 10^{-18} {~J} \\
&E_{\text {rad }}=h_{v} ; \quad v=\frac{E}{h}=\frac{2.95 \times 10^{-18}}{6.63 \times 10^{-34}}= \boxed{4.45e15} {~s}^{-1}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 172 |
In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{CO}$? | \boxed{1}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 173 |
Preamble: Electrons are accelerated by a potential of 10 Volts.
Subproblem 0: Determine their velocity in m/s. Please format your answer as $n \times 10^x$, where $n$ is to 2 decimal places.
Solution: The definition of an ${eV}$ is the energy gained by an electron when it is accelerated through a potential of $1 {~V}$, so an electron accelerated by a potential of $10 {~V}$ would have an energy of $10 {eV}$.\\
${E}=\frac{1}{2} m {v}^{2} \rightarrow {v}=\sqrt{2 {E} / {m}}$
\[
E=10 {eV}=1.60 \times 10^{-18} {~J}
\]
\[
\begin{aligned}
& {m}=\text { mass of electron }=9.11 \times 10^{-31} {~kg} \\
& v=\sqrt{\frac{2 \times 1.6 \times 10^{-18} {~J}}{9.11 \times 10^{-31} {~kg}}}= \boxed{1.87e6} {~m} / {s}
\end{aligned}
\]
Final answer: The final answer is 1.87e6. I hope it is correct.
Subproblem 1: Determine their deBroglie wavelength $\left(\lambda_{p}\right)$ in m. Please format your answer as $n \times 10^x$, where $n$ is to 2 decimal places. | $\lambda_{p}=h / m v$
\[
\lambda_{p}=\frac{6.63 \times 10^{-34}}{9.11 \times 10^{-34} {~kg} \times 1.87 \times 10^{6} {~m} / {s}}= \boxed{3.89e-10} {~m}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 174 |
Preamble: In all likelihood, the Soviet Union and the United States together in the past exploded about ten hydrogen devices underground per year.
If each explosion converted about $10 \mathrm{~g}$ of matter into an equivalent amount of energy (a conservative estimate), how many $k J$ of energy were released per device? Please format your answer as $n \times 10^{x}$. | $\Delta \mathrm{E}=\Delta \mathrm{mc}^{2}=10 \mathrm{~g} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} \times\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)^{2}$ $=9 \times 10^{14} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}=9 \times 10^{14} \mathrm{~J}= \boxed{9e11} \mathrm{~kJ} /$ bomb. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 175 |
Preamble: Calculate the molecular weight in g/mole (to 2 decimal places) of each of the substances listed below.
Subproblem 0: $\mathrm{NH}_{4} \mathrm{OH}$
Solution: $\mathrm{NH}_{4} \mathrm{OH}$ :
$5 \times 1.01=5.05(\mathrm{H})$
$1 \times 14.01=14.01(\mathrm{~N})$
$1 \times 16.00=16.00(\mathrm{O})$
$\mathrm{NH}_{4} \mathrm{OH}= \boxed{35.06}$ g/mole
Final answer: The final answer is 35.06. I hope it is correct.
Subproblem 1: $\mathrm{NaHCO}_{3}$
Solution: $\mathrm{NaHCO}_{3}: 3 \times 16.00=48.00(\mathrm{O})$
$1 \times 22.99=22.99(\mathrm{Na})$
$1 \times 1.01=1.01$ (H)
$1 \times 12.01=12.01$ (C)
$\mathrm{NaHCO}_{3}= \boxed{84.01}$ g/mole
Final answer: The final answer is 84.01. I hope it is correct.
Subproblem 2: $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}: \quad 2 \times 12.01=24.02$ (C)
$6 \times 1.01=6.06(\mathrm{H})$
$1 \times 16.00=16.00(\mathrm{O})$
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}: \boxed{46.08}$ g/mole | Introduction to Solid State Chemistry (3.091 Fall 2010) | 176 |
Subproblem 0: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{CO}$?
Solution: \boxed{1}.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{O}_{2}$ (in decimal form)?
Solution: \boxed{0.5}.
Final answer: The final answer is 0.5. I hope it is correct.
Subproblem 2: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{CO}_{2}$ (in decimal form)?
Solution: \boxed{1}.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 3: If $32.0 \mathrm{~g}$ of oxygen react with $\mathrm{CO}$ to form carbon dioxide $\left(\mathrm{CO}_{2}\right)$, how much CO was consumed in this reaction (to 1 decimal place)? | Molecular Weight (M.W.) of (M.W.) of $\mathrm{O}_{2}: 32.0$
(M.W.) of CO: $28.0$
available oxygen: $32.0 \mathrm{~g}=1$ mole, correspondingly the reaction involves 2 moles of CO [see (a)]:
\[
\mathrm{O}_{2}+2 \mathrm{CO} \rightarrow 2 \mathrm{CO}_{2}
\]
mass of CO reacted $=2$ moles $\times 28 \mathrm{~g} /$ mole $=\boxed{56.0} g$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 177 |
Preamble: For "yellow radiation" (frequency, $v,=5.09 \times 10^{14} s^{-1}$ ) emitted by activated sodium, determine:
the wavelength $(\lambda)$ in m. Please format your answer as $n \times 10^x$, where n is to 2 decimal places. | The equation relating $v$ and $\lambda$ is $c=v \lambda$ where $c$ is the speed of light $=3.00 \times 10^{8} \mathrm{~m}$.
\[
\lambda=\frac{c}{v}=\frac{3.00 \times 10^{8} m / s}{5.09 \times 10^{14} s^{-1}}=\boxed{5.89e-7} m
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 178 |
For a proton which has been subjected to an accelerating potential (V) of 15 Volts, determine its deBroglie wavelength in m. Please format your answer as $n \times 10^x$, where $n$ is to 1 decimal place. | \[
\begin{gathered}
E_{{K}}={eV}=\frac{{m}_{{p}} {v}^{2}}{2} ; \quad {v}_{{p}}=\sqrt{\frac{2 {eV}}{{m}_{{p}}}} \\
\lambda_{{p}}=\frac{{h}}{{m}_{{p}} {v}}=\frac{{h}}{{m}_{{p}} \sqrt{\frac{2 {eV}}{{m}_{{p}}}}}=\frac{{h}}{\sqrt{2 {eVm_{p }}}}=\frac{6.63 \times 10^{-34}}{\left(2 \times 1.6 \times 10^{-19} \times 15 \times 1.67 \times 10^{-27}\right)^{\frac{1}{2}}}
\\
= \boxed{7.4e-12} {~m}
\end{gathered}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 179 |
Preamble: For light with a wavelength $(\lambda)$ of $408 \mathrm{~nm}$ determine:
the frequency in $s^{-1}$. Please format your answer as $n \times 10^x$, where $n$ is to 3 decimal places. | To solve this problem we must know the following relationships:
\[
\begin{aligned}
v \lambda &=c
\end{aligned}
\]
$v$ (frequency) $=\frac{c}{\lambda}=\frac{3 \times 10^{8} m / s}{408 \times 10^{-9} m}= \boxed{7.353e14} s^{-1}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 180 |
Determine in units of eV (to 2 decimal places) the energy of a photon ( $h v)$ with the wavelength of $800$ nm. | \[
\begin{aligned}
E_{(\mathrm{eV})}=\frac{\mathrm{hc}}{\lambda} \times \frac{\mathrm{leV}}{1.6 \times 10^{-19} \mathrm{~J}} &=\frac{6.63 \times 10^{-34}[\mathrm{~s}] \times 3 \times 10^{8}\left[\frac{\mathrm{m}}{\mathrm{s}}\right]}{8.00 \times 10^{-7} \mathrm{~m}} \times \frac{\mathrm{leV}}{1.6 \times 10^{-19} \mathrm{~J}} \\
=\boxed{1.55} eV
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 181 |
Determine for barium (Ba) the linear density of atoms along the $<110>$ directions, in atoms/m. | Determine the lattice parameter and look at the unit cell occupation.
\includegraphics[scale=0.5]{set_23_img_02.jpg}
\nonessentialimage
Ba: $\quad$ BCC; atomic volume $=39.24 \mathrm{~cm}^{3} / \mathrm{mole} ; \mathrm{n}=2 \mathrm{atoms} /$ unit cell\\
$$
3.924 \times 10^{-5}\left(\mathrm{~m}^{3} / \text { mole }\right)=\frac{\mathrm{N}_{\mathrm{A}}}{2} \mathrm{a}^{3}
$$
$$
a=\sqrt[3]{\frac{2 \times 3.924 \times 10^{-5}}{6.02 \times 10^{23}}}=5.08 \times 10^{-10} \mathrm{~m}
$$
$$
\text { linear density }=\frac{1 \text { atom }}{a \sqrt{2}}=\frac{1}{5.08 \times 10^{-10} \times \sqrt{2}} = \boxed{1.39e9}
$$ atoms/m | Introduction to Solid State Chemistry (3.091 Fall 2010) | 182 |
A photon with a wavelength $(\lambda)$ of $3.091 \times 10^{-7} {~m}$ strikes an atom of hydrogen. Determine the velocity in m/s of an electron ejected from the excited state, $n=3$. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | \[
\begin{aligned}
&E_{\text {incident photon }}=E_{\text {binding }}+E_{\text {scattered } e^{-}} \\
&E_{\text {binding }}=-K\left(\frac{1}{3^{2}}\right) \quad \therefore \frac{hc}{\lambda}=\frac{K}{9}+\frac{1}{2} {mv^{2 }} \quad \therefore\left[\left(\frac{{hc}}{\lambda}-\frac{{K}}{9}\right) \frac{2}{{m}}\right]^{\frac{1}{2}}={v} \\
&{E}_{\text {incident photon }}=\frac{{hc}}{\lambda}=\frac{1}{2} {mv}^{2} \\
&{\left[\left(\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{3.091 \times 10^{-7}}-\frac{2.18 \times 10^{-18}}{9}\right) \frac{2}{9.11 \times 10^{-31}}\right]^{\frac{1}{2}}={v}} \\
&\therefore {v}= \boxed{9.35e5} {m} / {s}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 183 |
Preamble: For the element copper (Cu) determine:
the distance of second nearest neighbors (in meters). Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | The answer can be found by looking at a unit cell of $\mathrm{Cu}$ (FCC).
\includegraphics[scale=0.5]{set_23_img_00.jpg}
\nonessentialimage
Nearest neighbor distance is observed along $<110>$; second-nearest along $<100>$. The second-nearest neighbor distance is found to be "a".
Cu: atomic volume $=7.1 \times 10^{-6} \mathrm{~m}^{3} /$ mole $=\frac{\mathrm{N}_{\mathrm{A}}}{4} \mathrm{a}^{3}$ ( $\mathrm{Cu}: \mathrm{FCC} ; 4$ atoms/unit cell) $a=\sqrt[3]{\frac{7.1 \times 10^{-6} \times 4}{6.02 \times 10^{23}}}= \boxed{3.61e-10} \mathrm{~m}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 184 |
A line of the Lyman series of the spectrum of hydrogen has a wavelength of $9.50 \times 10^{-8} {~m}$. What was the "upper" quantum state $\left({n}_{{i}}\right)$ involved in the associated electron transition? | The Lyman series in hydrogen spectra comprises all electron transitions terminating in the ground state $({n}=1)$. In the present problem it is convenient to convert $\lambda$ into $\bar{v}$ and to use the Rydberg equation. Since we have an "emission spectrum", the sign will be negative in the conventional approach. We can avoid the sign problem, however:
\[
\begin{aligned}
& \bar{v}=R\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)=R\left(1-\frac{1}{n_{i}^{2}}\right) \\
& \overline{\frac{v}{R}}=\left(1-\frac{1}{n_{i}^{2}}\right) \\
& \frac{1}{n_{i}^{2}}=1-\frac{\bar{v}}{R}=\frac{R-\bar{v}}{R} \\
& n_{i}^{2}=\frac{R}{R-\bar{v}} \\
& {n}_{{i}}^{2}=\sqrt{\frac{{R}}{{R}-\bar{v}}} \quad \bar{v}=\frac{1}{9.5 \times 10^{-8} {~m}}=1.053 \times 10^{7} {~m}^{-1} \\
& n_{i}=\sqrt{\frac{1.097 \times 10^{7}}{1.097 \times 10^{7}-1.053 \times 10^{7}}}= \boxed{5}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 185 |
Determine the diffusivity $\mathrm{D}$ of lithium ( $\mathrm{Li}$ ) in silicon (Si) at $1200^{\circ} \mathrm{C}$, knowing that $D_{1100^{\circ} \mathrm{C}}=10^{-5} \mathrm{~cm}^{2} / \mathrm{s}$ and $\mathrm{D}_{695^{\circ} \mathrm{C}}=10^{-6} \mathrm{~cm}^{2} / \mathrm{s}$. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places, in $\mathrm{~cm}^2/\mathrm{sec}$. | \[
\begin{aligned}
&\frac{D_{1}}{D_{2}}=\frac{10^{-6}}{10^{-5}}=10^{-1}=e^{-\frac{E_{A}}{R}\left(\frac{1}{968}-\frac{1}{1373}\right)} \\
&E_{A}=\frac{R \ln 10}{\frac{1}{968}-\frac{1}{1373}}=62.8 \mathrm{~kJ} / \mathrm{mole} \\
&\frac{D_{1100}}{D_{1200}}=e^{-\frac{E_{A}}{R}\left(\frac{1}{1373}-\frac{1}{1473}\right)} \\
&D_{1200}=10^{-5} \times e^{\frac{E_{A}}{R}\left(\frac{1}{1373}-\frac{1}{1473}\right)}= \boxed{1.45e-5} \mathrm{~cm}^{2} / \mathrm{sec}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 186 |
By planar diffusion of antimony (Sb) into p-type germanium (Ge), a p-n junction is obtained at a depth of $3 \times 10^{-3} \mathrm{~cm}$ below the surface. What is the donor concentration in the bulk germanium if diffusion is carried out for three hours at $790^{\circ} \mathrm{C}$? Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places, and express it in units of $1/\mathrm{cm}^3$. The surface concentration of antimony is held constant at a value of $8 \times 10^{18}$ $\mathrm{cm}^{-3} ; D_{790^{\circ} \mathrm{C}}=4.8 \times 10^{-11} \mathrm{~cm}^{2} / \mathrm{s}$. | \includegraphics[scale=0.5]{set_37_img_00.jpg}
\nonessentialimage
\[
\begin{aligned}
&\frac{c}{c_{s}}=\operatorname{erfc} \frac{x}{2 \sqrt{D t}}=\operatorname{erfc} \frac{3 \times 10^{-3}}{2 \sqrt{D t}}=\operatorname{erfc}(2.083) \\
&\frac{c}{c_{s}}=1-\operatorname{erf}(2.083), \therefore 1-\frac{c}{c_{s}}=0.9964 \\
&\frac{c}{c_{s}}=3.6 \times 10^{-3}, \therefore c=2.88 \times 10^{16} \mathrm{~cm}^{-3}
\end{aligned}
\]
The donor concentration in germanium is $\boxed{2.88e16} / \mathrm{cm}^{3}$. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 187 |
Preamble: One mole of electromagnetic radiation (light, consisting of energy packages called photons) has an energy of $171 \mathrm{~kJ} /$ mole photons.
Determine the wavelength of this light in nm. | We know: $E_{\text {photon }}=h v=h c / \lambda$ to determine the wavelength associated with a photon we need to know its energy. $E=\frac{171 \mathrm{~kJ}}{\text { mole }}=\frac{1.71 \times 10^{5} \mathrm{~J}}{\text { mole }} \times \frac{1 \text { mole }}{6.02 \times 10^{23} \text { photons }}$
\[
=\frac{2.84 \times 10^{-19} \mathrm{~J}}{\text { photon }} ; \quad \mathrm{E}_{\text {photon }}=2.84 \times 10^{-19} \mathrm{~J}=\mathrm{h}_{v}=\frac{\mathrm{hc}}{\lambda}
\]
\[
\begin{aligned}
& \lambda=\frac{h c}{E_{\text {photon }}}=\frac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}}}{2.84 \times 10^{-19} \mathrm{~J}}=7.00 \times 10^{-7} \mathrm{~m} \\
& =\boxed{700} nm
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 188 |
Preamble: Two lasers generate radiation of (1) $9.5 \mu {m}$ and (2) $0.1 \mu {m}$ respectively.
Determine the photon energy (in eV, to two decimal places) of the laser generating radiation of $9.5 \mu {m}$. | \[
\begin{aligned}
{E} &={h} v=\frac{{hc}}{\lambda} {J} \times \frac{1 {eV}}{1.6 \times 10^{-19} {~J}} \\
{E}_{1} &=\frac{{hc}}{9.5 \times 10^{-6}} \times \frac{1}{1.6 \times 10^{-19}} {eV}= \boxed{0.13} {eV}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 189 |
At $100^{\circ} \mathrm{C}$ copper $(\mathrm{Cu})$ has a lattice constant of $3.655 \AA$. What is its density in $g/cm^3$ at this temperature? Please round your answer to 2 decimal places. | $\mathrm{Cu}$ is FCC, so $\mathrm{n}=4$
\[
\begin{aligned}
&\mathrm{a}=3.655 \AA=3.655 \times 10^{-10} \mathrm{~m} \\
&\text { atomic weight }=63.55 \mathrm{~g} / \mathrm{mole} \\
&\frac{\text { atomic weight }}{\rho} \times 10^{-6}=\frac{N_{\mathrm{A}}}{\mathrm{n}} \times \mathrm{a}^{3} \\
&\rho=\frac{(63.55 \mathrm{~g} / \mathrm{mole})(4 \text { atoms } / \text { unit cell })}{\left(6.023 \times 10^{23} \text { atoms } / \mathrm{mole}\right)\left(3.655 \times 10^{-10} \mathrm{~m}^{3}\right)}= \boxed{8.64} \mathrm{~g} / \mathrm{cm}^{3}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 190 |
Determine the atomic (metallic) radius of Mo in meters. Do not give the value listed in the periodic table; calculate it from the fact that Mo's atomic weight is $=95.94 \mathrm{~g} /$ mole and $\rho=10.2 \mathrm{~g} / \mathrm{cm}^{3}$. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | Mo: atomic weight $=95.94 \mathrm{~g} /$ mole
\[
\rho=10.2 \mathrm{~g} / \mathrm{cm}^{3}
\]
BCC, so $n=2$ atoms/unit cell
\[
\begin{aligned}
&\mathrm{a}^{3}=\frac{(95.94 \mathrm{~g} / \mathrm{mole})(2 \text { atoms/unit cell })}{\left(10.2 \mathrm{~g} / \mathrm{cm}^{3}\right)\left(6.023 \times 10^{23} \text { atoms } / \mathrm{mole}\right)} \times 10^{-6} \frac{\mathrm{m}^{3}}{\mathrm{~cm}^{3}} \\
&=3.12 \times 10^{-29} \mathrm{~m}^{3} \\
&a=3.22 \times 10^{-10} \mathrm{~m}
\end{aligned}
\]
For BCC, $a \sqrt{3}=4 r$, so $r= \boxed{1.39e-10} \mathrm{~m}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 191 |
Preamble: Determine the following values from a standard radio dial.
What is the minimum wavelength in m for broadcasts on the AM band? Format your answer as an integer. | \[
\mathrm{c}=v \lambda, \therefore \lambda_{\min }=\frac{\mathrm{c}}{v_{\max }} ; \lambda_{\max }=\frac{\mathrm{c}}{v_{\min }}
\]
$\lambda_{\min }=\frac{3 \times 10^{8} m / s}{1600 \times 10^{3} Hz}=\boxed{188} m$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 192 |
Consider a (111) plane in an FCC structure. How many different [110]-type directions lie in this (111) plane? | Let's look at the unit cell.
\includegraphics[scale=0.5]{set_23_img_01.jpg}
\nonessentialimage
There are \boxed{6} [110]-type directions in the (111) plane. Their indices are:
\[
(10 \overline{1}),(\overline{1} 01),(\overline{1} 10),(\overline{1} 0),(0 \overline{1} 1),(01 \overline{1})
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 193 |
Determine the velocity of an electron (in $\mathrm{m} / \mathrm{s}$ ) that has been subjected to an accelerating potential $V$ of 150 Volt. Please format your answer as $n \times 10^x$, where $n$ is to 2 decimal places.
(The energy imparted to an electron by an accelerating potential of one Volt is $1.6 \times 10^{-19}$ J oules; dimensional analysis shows that the dimensions of charge $x$ potential correspond to those of energy; thus: 1 electron Volt $(1 \mathrm{eV})=1.6 \times 10^{-19}$ Coulomb $\times 1$ Volt $=1.6 \times 10^{-19}$ Joules.) | We know: $E_{\text {kin }}=m v^{2} / 2=e \times V$ (charge applied potential) $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}$
\[
\begin{aligned}
&E_{\text {kin }}=e \times V=m v^{2} / 2 \\
&v=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 150}{9.1 \times 10^{-31}}}=\boxed{7.26e6} \mathrm{~m} / \mathrm{s}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 194 |
In a diffractometer experiment a specimen of thorium (Th) is irradiated with tungsten (W) $L_{\alpha}$ radiation. Calculate the angle, $\theta$, of the $4^{\text {th }}$ reflection. Round your answer (in degrees) to 2 decimal places. | $\bar{v}=\frac{1}{\lambda}=\frac{5}{36}(74-7.4)^{2} \mathrm{R} \rightarrow \lambda=1.476 \times 10^{-10} \mathrm{~m}$
Th is FCC with a value of $\mathrm{V}_{\text {molar }}=19.9 \mathrm{~cm}^{3}$
$\therefore \frac{4}{\mathrm{a}^{3}}=\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{V}_{\text {molar }}} \rightarrow \mathrm{a}=\left(\frac{4 \times 19.9}{6.02 \times 10^{23}}\right)^{1 / 3}=5.095 \times 10^{-8} \mathrm{~cm}$
$\lambda=2 d \sin \theta ; d=\frac{a}{\sqrt{h^{2}+k^{2}+R^{2}}}$
4th reflection in FCC: $111 ; 200 ; 220 ; \mathbf{3 1 1} \rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{l}^{2}=11$
$\lambda_{\theta}=\frac{2 a \sin \theta}{\sqrt{h^{2}+k^{2}+L^{2}}} \rightarrow=\sin ^{-1}\left(\frac{\lambda \sqrt{h^{2}+k^{2}+l^{2}}}{2 a}\right)=\sin ^{-1}\left(\frac{1.476 \sqrt{11}}{2 \times 5.095}\right)=\boxed{28.71}^{\circ}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 195 |
A metal is found to have BCC structure, a lattice constant of $3.31 \AA$, and a density of $16.6 \mathrm{~g} / \mathrm{cm}^{3}$. Determine the atomic weight of this element in g/mole, and round your answer to 1 decimal place. | $B C C$ structure, so $\mathrm{n}=2$
\[
\begin{aligned}
&a=3.31 \AA=3.31 \times 10^{-10} \mathrm{~m} \\
&\rho=16.6 \mathrm{~g} / \mathrm{cm}^{3} \\
&\frac{\text { atomic weight }}{\rho} \times 10^{-6}=\frac{N_{A}}{n} \times a^{3}
\end{aligned}
\]
\[
\begin{aligned}
&\text { atomic weight }=\frac{\left(6.023 \times 10^{23} \text { atoms } / \text { mole }\right)\left(3.31 \times 10^{-10} \mathrm{~m}\right)^{3}}{(2 \text { atoms } / \text { unit cell })\left(10^{-6} \mathrm{~m}^{3} / \mathrm{cm}^{3}\right)} \times 16.6 \mathrm{~g} / \mathrm{cm}^{3} \\
&= \boxed{181.3} \mathrm{~g} / \text { mole }
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 196 |
Preamble: Iron $\left(\rho=7.86 \mathrm{~g} / \mathrm{cm}^{3}\right.$ ) crystallizes in a BCC unit cell at room temperature.
Calculate the radius in cm of an iron atom in this crystal. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | In $\mathrm{BCC}$ there are 2 atoms per unit cell, so $\frac{2}{\mathrm{a}^{3}}=\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{V}_{\text {molar }}}$, where $\mathrm{V}_{\text {molar }}=\mathrm{A} / \rho ; \mathrm{A}$ is the atomic mass of iron.
\[
\begin{aligned}
&\frac{2}{a^{3}}=\frac{N_{A} \times p}{A} \\
&\therefore a=\left(\frac{2 A}{N_{A} \times \rho}\right)^{\frac{1}{3}}=\frac{4}{\sqrt{3}} r \\
&\therefore r= \boxed{1.24e-8} \mathrm{~cm}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 197 |
Preamble: For the element copper (Cu) determine:
Subproblem 0: the distance of second nearest neighbors (in meters). Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places.
Solution: The answer can be found by looking at a unit cell of $\mathrm{Cu}$ (FCC).
\includegraphics[scale=0.5]{set_23_img_00.jpg}
\nonessentialimage
Nearest neighbor distance is observed along $<110>$; second-nearest along $<100>$. The second-nearest neighbor distance is found to be "a".
Cu: atomic volume $=7.1 \times 10^{-6} \mathrm{~m}^{3} /$ mole $=\frac{\mathrm{N}_{\mathrm{A}}}{4} \mathrm{a}^{3}$ ( $\mathrm{Cu}: \mathrm{FCC} ; 4$ atoms/unit cell) $a=\sqrt[3]{\frac{7.1 \times 10^{-6} \times 4}{6.02 \times 10^{23}}}= \boxed{3.61e-10} \mathrm{~m}$
Final answer: The final answer is 3.61e-10. I hope it is correct.
Subproblem 1: the interplanar spacing of $\{110\}$ planes (in meters). Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | $d_{h k l}=\frac{a}{\sqrt{h^{2}+k^{2}+1^{2}}}$
\[
d_{110}=\frac{3.61 \times 10^{-10}}{\sqrt{2}}= \boxed{2.55e-10} \mathrm{~m}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 198 |
Subproblem 0: What is the working temperature for silica glass in Celsius?
Solution: \boxed{1950}.
Final answer: The final answer is 1950. I hope it is correct.
Subproblem 1: What is the softening temperature for silica glass in Celsius?
Solution: \boxed{1700}.
Final answer: The final answer is 1700. I hope it is correct.
Subproblem 2: What is the working temperature for Pyrex in Celsius?
Solution: \boxed{1200}.
Final answer: The final answer is 1200. I hope it is correct.
Subproblem 3: What is the softening temperature for Pyrex in Celsius?
Solution: \boxed{800}.
Final answer: The final answer is 800. I hope it is correct.
Subproblem 4: What is the working temperature for soda-lime glass in Celsius? | \boxed{900}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 199 |
Subsets and Splits