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What is the total charge in coulombs of $75.0 \mathrm{~kg}$ of electrons? | $-1.32$ | -1.32 | Question 21.59 | $10^{13} \mathrm{C}$ | fund |
||
A uniformly charged conducting sphere of $1.2 \mathrm{~m}$ diameter has surface charge density $8.1 \mu \mathrm{C} / \mathrm{m}^2$. Find the net charge on the sphere. | $37$ | 37 | Question 23.17 | $\mu \mathrm{C}$ | fund |
||
The magnitude of the electrostatic force between two identical ions that are separated by a distance of $5.0 \times 10^{-10} \mathrm{~m}$ is $3.7 \times 10^{-9}$ N. What is the charge of each ion? | $3.2$ | 3.2 | Question 21.27 | $10^{-19} \mathrm{C}$ | fund |
||
How many megacoulombs of positive charge are in $1.00 \mathrm{~mol}$ of neutral molecular-hydrogen gas $\left(\mathrm{H}_2\right)$ ? | $0.19$ | 0.19 | Question 21.45 | $\mathrm{MC}$ | fund |
||
A charge (uniform linear density $=9.0 \mathrm{nC} / \mathrm{m}$ ) lies on a string that is stretched along an $x$ axis from $x=0$ to $x=3.0 \mathrm{~m}$. Determine the magnitude of the electric field at $x=4.0 \mathrm{~m}$ on the $x$ axis. | $61$ | 61 | Question 22.67 | $\mathrm{~N} / \mathrm{C}$ | fund |
||
A long, straight wire has fixed negative charge with a linear charge density of magnitude $3.6 \mathrm{nC} / \mathrm{m}$. The wire is to be enclosed by a coaxial, thin-walled nonconducting cylindrical shell of radius $1.5 \mathrm{~cm}$. The shell is to have positive charge on its outside surface with a surface charge density $\sigma$ that makes the net external electric field zero. Calculate $\sigma$.
| $3.8$ | 3.8 | Question 23.27 | $10^{-8} \mathrm{C} / \mathrm{m}^2$ | fund |
||
Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. What acceleration would a proton experience if the gun's electric field were $2.00 \times 10^4 \mathrm{~N} / \mathrm{C}$ ? | $1.92$ | 1.92 | Question 22.47 | $10^{12} \mathrm{~m} / \mathrm{s}^2 $ | fund |
||
An infinite line of charge produces a field of magnitude $4.5 \times$ $10^4 \mathrm{~N} / \mathrm{C}$ at distance $2.0 \mathrm{~m}$. Find the linear charge density.
| $5.0$ | 5.0 | Question 23.25 | $\mu \mathrm{C} / \mathrm{m}$ | fund |
||
A charged nonconducting rod, with a length of $2.00 \mathrm{~m}$ and a cross-sectional area of $4.00 \mathrm{~cm}^2$, lies along the positive side of an $x$ axis with one end at the origin. The volume charge density $\rho$ is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if $\rho$ is uniform, with a value of $-4.00 \mu \mathrm{C} / \mathrm{m}^3$?
| $2.00$ | 2.00 | Question 21.51 | $10^{10} \text { electrons; }$ | fund |
||
At $x=6.5 \mathrm{~m}$, the particle has kinetic energy
$$
\begin{aligned}
K_0 & =\frac{1}{2} m v_0^2=\frac{1}{2}(2.00 \mathrm{~kg})(4.00 \mathrm{~m} / \mathrm{s})^2 \\
& =16.0 \mathrm{~J} .
\end{aligned}
$$
Because the potential energy there is $U=0$, the mechanical energy is
$$
E_{\text {mec }}=K_0+U_0=16.0 \mathrm{~J}+0=16.0 \mathrm{~J} .
$$The kinetic energy $K_1$ is the difference between $E_{\text {mec }}$ and $U_1$ :
$$
K_1=E_{\text {mec }}-U_1=16.0 \mathrm{~J}-7.0 \mathrm{~J}=9.0 \mathrm{~J} .
$$
Because $K_1=\frac{1}{2} m v_1^2$, we find
$$
v_1=3.0 \mathrm{~m} / \mathrm{s}
$$
| A $2.00 \mathrm{~kg}$ particle moves along an $x$ axis in one-dimensional motion while a conservative force along that axis acts on it. The potential energy $U(x)$ is 0 when $x = 6.5 \mathrm{~m} $ and is $7 \mathrm{~J}$ when $x = 4.5 \mathrm{~m} $. At $x=6.5 \mathrm{~m}$, the particle has velocity $\vec{v}_0=(-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}$ Determine the particle's speed at $x_1=4.5 \mathrm{~m}$. | 3.0 | 3.0 | 8.04 | m/s | fund |
|
We can get $E$ from the orbital energy, given by Eq. $13-40$ ( $E=-G M m / 2 r)$ ), if we first find the orbital radius $r$. (It is not simply the given altitude.)
Calculations: The orbital radius must be
$$
r=R+h=6370 \mathrm{~km}+350 \mathrm{~km}=6.72 \times 10^6 \mathrm{~m},
$$
in which $R$ is the radius of Earth. Then, from Eq. 13-40 with Earth mass $M=5.98 \times 10^{24} \mathrm{~kg}$, the mechanical energy is
$$
\begin{aligned}
E & =-\frac{G M m}{2 r} \\
& =-\frac{\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{kg}^2\right)\left(5.98 \times 10^{24} \mathrm{~kg}\right)(7.20 \mathrm{~kg})}{(2)\left(6.72 \times 10^6 \mathrm{~m}\right)} \\
& =-2.14 \times 10^8 \mathrm{~J}=-214 \mathrm{MJ} .
\end{aligned}
$$ | A playful astronaut releases a bowling ball, of mass $m=$ $7.20 \mathrm{~kg}$, into circular orbit about Earth at an altitude $h$ of $350 \mathrm{~km}$.
What is the mechanical energy $E$ of the ball in its orbit? | -214 | -214 | 13.05 | MJ | fund |
|
We can find $K$ with Eq. $10-34\left(K=\frac{1}{2} I \omega^2\right)$. We already know that $I=\frac{1}{2} M R^2$, but we do not yet know $\omega$ at $t=2.5 \mathrm{~s}$. However, because the angular acceleration $\alpha$ has the constant value of $-24 \mathrm{rad} / \mathrm{s}^2$, we can apply the equations for constant angular acceleration.
Calculations: Because we want $\omega$ and know $\alpha$ and $\omega_0(=0)$, we use Eq. 10-12:
$$
\omega=\omega_0+\alpha t=0+\alpha t=\alpha t .
$$
Substituting $\omega=\alpha t$ and $I=\frac{1}{2} M R^2$ into Eq. 10-34, we find
$$
\begin{aligned}
K & =\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{1}{2} M R^2\right)(\alpha t)^2=\frac{1}{4} M(R \alpha t)^2 \\
& =\frac{1}{4}(2.5 \mathrm{~kg})\left[(0.20 \mathrm{~m})\left(-24 \mathrm{rad} / \mathrm{s}^2\right)(2.5 \mathrm{~s})\right]^2 \\
& =90 \mathrm{~J} .
\end{aligned}
$$
| There is a disk mounted on a fixed horizontal axle. A block hangs from a massless cord that is wrapped around the rim of the disk. The cord does not slip and there is no friction at the axle. Let the disk start from rest at time $t=0$ and also let the tension in the massless cord be $6.0 \mathrm{~N}$ and the angular acceleration of the disk be $-24 \mathrm{rad} / \mathrm{s}^2$. What is its rotational kinetic energy $K$ at $t=2.5 \mathrm{~s}$ ? | 90 | 90 | 10.11 | J | fund |
|
We can relate $\Delta E_{\text{th }}$ to the work $W$ done by $\vec{F}$ with the energy statement for a system that involves friction:
$$
W=\Delta E_{\text {mec }}+\Delta E_{\text {th }} .
$$
Calculations: We know the value of $W=F d \cos \phi=(40 \mathrm{~N})(0.50 \mathrm{~m}) \cos 0^{\circ}=20 \mathrm{~J}$. The change $\Delta E_{\text {mec }}$ in the crate's mechanical energy is just the change in its kinetic energy because no potential energy changes occur, so we have
$$
\Delta E_{\mathrm{mec}}=\Delta K=\frac{1}{2} m v^2-\frac{1}{2} m v_0^2 .
$$
Substituting this into above equation and solving for $\Delta E_{\mathrm{th}}$, we find
$$
\begin{aligned}
\Delta E_{\mathrm{th}} & =W-\left(\frac{1}{2} m v^2-\frac{1}{2} m v_0^2\right)=W-\frac{1}{2} m\left(v^2-v_0^2\right) \\
& =20 \mathrm{~J}-\frac{1}{2}(14 \mathrm{~kg})\left[(0.20 \mathrm{~m} / \mathrm{s})^2-(0.60 \mathrm{~m} / \mathrm{s})^2\right] \\
& =22.2 \mathrm{~J}.
\end{aligned}
$$
| A food shipper pushes a wood crate of cabbage heads (total mass $m=14 \mathrm{~kg}$ ) across a concrete floor with a constant horizontal force $\vec{F}$ of magnitude $40 \mathrm{~N}$. In a straight-line displacement of magnitude $d=0.50 \mathrm{~m}$, the speed of the crate decreases from $v_0=0.60 \mathrm{~m} / \mathrm{s}$ to $v=0.20 \mathrm{~m} / \mathrm{s}$. What is the increase $\Delta E_{\text {th }}$ in the thermal energy of the crate and floor? | 22.2 | 22.2 | 8.05 | J | fund |
|
Because the cylinder's angular acceleration is constant, we can relate it to the angular velocity and angular displacement via the basic equations for constant angular acceleration.
Calculations: Let's first do a quick check to see if we can solve the basic equations. The initial angular velocity is $\omega_0=3.40$
$\mathrm{rad} / \mathrm{s}$, the angular displacement is $\theta-\theta_0=20.0 \mathrm{rev}$, and the angular velocity at the end of that displacement is $\omega=2.00$ $\mathrm{rad} / \mathrm{s}$. In addition to the angular acceleration $\alpha$ that we want, both basic equations also contain time $t$, which we do not necessarily want.
To eliminate the unknown $t$, we use $$\omega =\omega_0+\alpha t$$ to write
$$
t=\frac{\omega-\omega_0}{\alpha}
$$
which we then substitute into equation $$\theta-\theta_0 =\omega_0 t+\frac{1}{2} \alpha t^2 $$ to write
$$
\theta-\theta_0=\omega_0\left(\frac{\omega-\omega_0}{\alpha}\right)+\frac{1}{2} \alpha\left(\frac{\omega-\omega_0}{\alpha}\right)^2 .
$$
Solving for $\alpha$, substituting known data, and converting 20 rev to $125.7 \mathrm{rad}$, we find
$$
\begin{aligned}
\alpha & =\frac{\omega^2-\omega_0^2}{2\left(\theta-\theta_0\right)}=\frac{(2.00 \mathrm{rad} / \mathrm{s})^2-(3.40 \mathrm{rad} / \mathrm{s})^2}{2(125.7 \mathrm{rad})} \\
& =-0.0301 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$$
| While you are operating a Rotor (a large, vertical, rotating cylinder found in amusement parks), you spot a passenger in acute distress and decrease the angular velocity of the cylinder from $3.40 \mathrm{rad} / \mathrm{s}$ to $2.00 \mathrm{rad} / \mathrm{s}$ in $20.0 \mathrm{rev}$, at constant angular acceleration. (The passenger is obviously more of a "translation person" than a "rotation person.")
What is the constant angular acceleration during this decrease in angular speed? | -0.0301 | -0.0301 | 10.04 | $\mathrm{rad} / \mathrm{s}^2$ | fund |
|
(1) The air's weight is equal to $m g$, where $m$ is its mass.
(2) Mass $m$ is related to the air density $\rho$ and the air volume $V$ by Eq $(\rho=m / V)$.
Calculation: Putting the two ideas together and taking the density of air at 1.0 atm, we find
$$
\begin{aligned}
m g & =(\rho V) g \\
& =\left(1.21 \mathrm{~kg} / \mathrm{m}^3\right)(3.5 \mathrm{~m} \times 4.2 \mathrm{~m} \times 2.4 \mathrm{~m})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right) \\
& =418 \mathrm{~N}
\end{aligned}
$$
| A living room has floor dimensions of $3.5 \mathrm{~m}$ and $4.2 \mathrm{~m}$ and a height of $2.4 \mathrm{~m}$.
What does the air in the room weigh when the air pressure is $1.0 \mathrm{~atm}$ ? | 418 | 418 | 14.01 | N | fund |
|
We can approximate Earth as a uniform sphere of mass $M_E$. Then the gravitational acceleration at any distance $r$ from the center of Earth is
$$
a_g=\frac{G M_E}{r^2}
$$
We might simply apply this equation twice, first with $r=$ $6.77 \times 10^6 \mathrm{~m}$ for the location of the feet and then with $r=6.77 \times 10^6 \mathrm{~m}+1.70 \mathrm{~m}$ for the location of the head. However, a calculator may give us the same value for $a_g$ twice, and thus a difference of zero, because $h$ is so much smaller than $r$. Here's a more promising approach: Because we have a differential change $d r$ in $r$ between the astronaut's feet and head, we should differentiate above equation with respect to $r$.
Calculations: The differentiation gives us
$$
d a_g=-2 \frac{G M_E}{r^3} d r
$$
where $d a_g$ is the differential change in the gravitational acceleration due to the differential change $d r$ in $r$. For the astronaut, $d r=h$ and $r=6.77 \times 10^6 \mathrm{~m}$. Substituting data into equation, we find
$$
\begin{aligned}
d a_g & =-2 \frac{\left(6.67 \times 10^{-11} \mathrm{~m}^3 / \mathrm{kg} \cdot \mathrm{s}^2\right)\left(5.98 \times 10^{24} \mathrm{~kg}\right)}{\left(6.77 \times 10^6 \mathrm{~m}\right)^3}(1.70 \mathrm{~m}) \\
& =-4.37 \times 10^{-6} \mathrm{~m} / \mathrm{s}^2, \quad \text { (Answer) }
\end{aligned}
$$
| An astronaut whose height $h$ is $1.70 \mathrm{~m}$ floats "feet down" in an orbiting space shuttle at distance $r=6.77 \times 10^6 \mathrm{~m}$ away from the center of Earth. What is the difference between the gravitational acceleration at her feet and at her head? | -4.37 | -4.37 | 13.02 | $10 ^ {-6}m/s^2$ | fund |
|
We can treat the center of mass as if it were a real particle, with a mass equal to the system's total mass $M=16 \mathrm{~kg}$. We can also treat the three external forces as if they act at the center of mass.
Calculations: We can now apply Newton's second law $\left(\vec{F}_{\text {net }}=m \vec{a}\right)$ to the center of mass, writing
$$
\vec{F}_{\text {net }}=M \vec{a}_{\mathrm{com}}
$$
or
$$
\begin{aligned}
& \vec{F}_1+\vec{F}_2+\vec{F}_3=M \vec{a}_{\mathrm{com}} \\
& \vec{a}_{\mathrm{com}}=\frac{\vec{F}_1+\vec{F}_2+\vec{F}_3}{M} .
\end{aligned}
$$
Equation tells us that the acceleration $\vec{a}_{\text {com }}$ of the center of mass is in the same direction as the net external force $\vec{F}_{\text {net }}$ on the system. Because the particles are initially at rest, the center of mass must also be at rest. As the center of mass then begins to accelerate, it must move off in the common direction of $\vec{a}_{\text {com }}$ and $\vec{F}_{\text {net }}$.
We can evaluate the right side of the above equation directly on a vector-capable calculator, or we can rewrite the equation in component form, find the components of $\vec{a}_{\text {com }}$, and then find $\vec{a}_{\text {com }}$. Along the $x$ axis, we have
$$
\begin{aligned}
a_{\mathrm{com}, x} & =\frac{F_{1 x}+F_{2 x}+F_{3 x}}{M} \\
& =\frac{-6.0 \mathrm{~N}+(12 \mathrm{~N}) \cos 45^{\circ}+14 \mathrm{~N}}{16 \mathrm{~kg}}=1.03 \mathrm{~m} / \mathrm{s}^2 .
\end{aligned}
$$
Along the $y$ axis, we have
$$
\begin{aligned}
a_{\mathrm{com}, y} & =\frac{F_{1 y}+F_{2 y}+F_{3 y}}{M} \\
& =\frac{0+(12 \mathrm{~N}) \sin 45^{\circ}+0}{16 \mathrm{~kg}}=0.530 \mathrm{~m} / \mathrm{s}^2 .
\end{aligned}
$$
From these components, we find that $\vec{a}_{\mathrm{com}}$ has the magnitude
$$
\begin{aligned}
a_{\mathrm{com}} & =\sqrt{\left(a_{\mathrm{com}, x}\right)^2+\left(a_{\text {com }, y}\right)^2} \\
& =1.16 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
| If the particles in a system all move together, the com moves with them-no trouble there. But what happens when they move in different directions with different accelerations? Here is an example.
The three particles in Figure are initially at rest. Each experiences an external force due to bodies outside the three-particle system. The directions are indicated, and the magnitudes are $F_1=6.0 \mathrm{~N}, F_2=12 \mathrm{~N}$, and $F_3=14 \mathrm{~N}$. What is the acceleration of the center of mass of the system? | 1.16 | 1.16 | 9.03 | $m/s^2$ | fund |
|
Because we are to neglect the effects of the atmosphere on the asteroid, the mechanical energy of the asteroid-Earth system is conserved during the fall. Thus, the final mechanical energy (when the asteroid reaches Earth's surface) is equal to the initial mechanical energy. With kinetic energy $K$ and gravitational potential energy $U$, we can write this as
$$
K_f+U_f=K_i+U_i
$$
Also, if we assume the system is isolated, the system's linear momentum must be conserved during the fall. Therefore, the momentum change of the asteroid and that of Earth must be equal in magnitude and opposite in sign. However, because Earth's mass is so much greater than the asteroid's mass, the change in Earth's speed is negligible relative to the change in the asteroid's speed. So, the change in Earth's kinetic energy is also negligible. Thus, we can assume that the kinetic energies in above equation are those of the asteroid alone.
Calculations: Let $m$ represent the asteroid's mass and $M$ represent Earth's mass $\left(5.98 \times 10^{24} \mathrm{~kg}\right)$. The asteroid is ini-
tially at distance $10 R_E$ and finally at distance $R_E$, where $R_E$ is Earth's radius $\left(6.37 \times 10^6 \mathrm{~m}\right)$. Substituting $-\frac{G M m}{r}$ for $U$ and $\frac{1}{2} m v^2$ for $K$, we rewrite above equation as
$$
\frac{1}{2} m v_f^2-\frac{G M m}{R_E}=\frac{1}{2} m v_i^2-\frac{G M m}{10 R_E} .
$$
Rearranging and substituting known values, we find
$$
\begin{aligned}
v_f^2= & v_i^2+\frac{2 G M}{R_E}\left(1-\frac{1}{10}\right) \\
= & \left(12 \times 10^3 \mathrm{~m} / \mathrm{s}\right)^2 \\
& +\frac{2\left(6.67 \times 10^{-11} \mathrm{~m}^3 / \mathrm{kg} \cdot \mathrm{s}^2\right)\left(5.98 \times 10^{24} \mathrm{~kg}\right)}{6.37 \times 10^6 \mathrm{~m}} 0.9 \\
= & 2.567 \times 10^8 \mathrm{~m}^2 / \mathrm{s}^2,
\end{aligned}
$$
and
$$
v_f=1.60 \times 10^4 \mathrm{~m} / \mathrm{s}
$$
| An asteroid, headed directly toward Earth, has a speed of $12 \mathrm{~km} / \mathrm{s}$ relative to the planet when the asteroid is 10 Earth radii from Earth's center. Neglecting the effects of Earth's atmosphere on the asteroid, find the asteroid's speed $v_f$ when it reaches Earth's surface. | 1.60 | 1.60 | 13.03 | $10^4 \mathrm{~m} / \mathrm{s}$ | fund |
|
Let the mechanical energy be $E_{\text {mec }, t}$ when the child is at the top of the slide and $E_{\mathrm{mec}, b}$ when she is at the bottom. Then the conservation principle tells us
$$
E_{\mathrm{mec}, b}=E_{\mathrm{mec}, t} .
$$
To show both kinds of mechanical energy, we have
$$
\begin{aligned}
K_b+U_b & =K_t+U_t, \\
\frac{1}{2} m v_b^2+m g y_b & =\frac{1}{2} m v_t^2+m g y_t .
\end{aligned}
$$
Dividing by $m$ and rearranging yield
$$
v_b^2=v_t^2+2 g\left(y_t-y_b\right)
$$
Putting $v_t=0$ and $y_t-y_b=h$ leads to
$$
\begin{aligned}
v_b & =\sqrt{2 g h}=\sqrt{(2)\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)(8.5 \mathrm{~m})} \\
& =13 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$
| The huge advantage of using the conservation of energy instead of Newton's laws of motion is that we can jump from the initial state to the final state without considering all the intermediate motion. Here is an example. In Figure, a child of mass $m$ is released from rest at the top of a water slide, at height $h=8.5 \mathrm{~m}$ above the bottom of the slide. Assuming that the slide is frictionless because of the water on it, find the child's speed at the bottom of the slide. | 13 | 13 | 8.03 | m/s | fund |
|
Radiation from an X-ray source consists of two components of wavelengths $154.433 \mathrm{pm}$ and $154.051 \mathrm{pm}$. Calculate the difference in glancing angles $(2 \theta)$ of the diffraction lines arising from the two components in a diffraction pattern from planes of separation $77.8 \mathrm{pm}$. | 2.14 | 2.14 | 37.7(a) | ${\circ}$ | matter |
||
A chemical reaction takes place in a container of cross-sectional area $50 \mathrm{~cm}^2$. As a result of the reaction, a piston is pushed out through $15 \mathrm{~cm}$ against an external pressure of $1.0 \mathrm{~atm}$. Calculate the work done by the system. | -75 | -75 | 55.4(a) | $\mathrm{~J}$ | matter |
||
A mixture of water and ethanol is prepared with a mole fraction of water of 0.60 . If a small change in the mixture composition results in an increase in the chemical potential of water by $0.25 \mathrm{~J} \mathrm{~mol}^{-1}$, by how much will the chemical potential of ethanol change? | -0.38 | -0.38 | 69.2(a) | $\mathrm{~J} \mathrm{~mol}^{-1}$ | matter |
||
The enthalpy of fusion of mercury is $2.292 \mathrm{~kJ} \mathrm{~mol}^{-1}$, and its normal freezing point is $234.3 \mathrm{~K}$ with a change in molar volume of $+0.517 \mathrm{~cm}^3 \mathrm{~mol}^{-1}$ on melting. At what temperature will the bottom of a column of mercury (density $13.6 \mathrm{~g} \mathrm{~cm}^{-3}$ ) of height $10.0 \mathrm{~m}$ be expected to freeze? | 234.4 | problem | 234.4 | 69.3 | $ \mathrm{~K}$ | matter |
|
Suppose a nanostructure is modelled by an electron confined to a rectangular region with sides of lengths $L_1=1.0 \mathrm{~nm}$ and $L_2=2.0 \mathrm{~nm}$ and is subjected to thermal motion with a typical energy equal to $k T$, where $k$ is Boltzmann's constant. How low should the temperature be for the thermal energy to be comparable to the zero-point energy？ | 5.5 | 5.5 | 11.3(a) | $10^3 \mathrm{~K}$ | matter |
||
Calculate the change in Gibbs energy of $35 \mathrm{~g}$ of ethanol (mass density $0.789 \mathrm{~g} \mathrm{~cm}^{-3}$ ) when the pressure is increased isothermally from 1 atm to 3000 atm. | 12 | 12 | 66.5(a) | $\mathrm{~kJ}$ | matter |
||
The promotion of an electron from the valence band into the conduction band in pure $\mathrm{TIO}_2$ by light absorption requires a wavelength of less than $350 \mathrm{~nm}$. Calculate the energy gap in electronvolts between the valence and conduction bands. | 3.54 | 3.54 | 39.1(a) | $\mathrm{eV}$ | matter |
||
Although the crystallization of large biological molecules may not be as readily accomplished as that of small molecules, their crystal lattices are no different. Tobacco seed globulin forms face-centred cubic crystals with unit cell dimension of $12.3 \mathrm{~nm}$ and a density of $1.287 \mathrm{~g} \mathrm{~cm}^{-3}$. Determine its molar mass. | 3.61 | problem | 3.61 | 37.1 | $10^5 \mathrm{~g} \mathrm{~mol}^{-1}$ | matter |
|
An electron is accelerated in an electron microscope from rest through a potential difference $\Delta \phi=100 \mathrm{kV}$ and acquires an energy of $e \Delta \phi$. What is its final speed? | 1.88 | 1.88 | 2.7(a) | $10^8 \mathrm{~m} \mathrm{~s}^{-1}$ | matter |
||
The following data show how the standard molar constant-pressure heat capacity of sulfur dioxide varies with temperature. By how much does the standard molar enthalpy of $\mathrm{SO}_2(\mathrm{~g})$ increase when the temperature is raised from $298.15 \mathrm{~K}$ to $1500 \mathrm{~K}$ ? | 62.2 | problem | 62.2 | 56.1 | $\mathrm{~kJ} \mathrm{~mol}^{-1}$ | matter |
|
Suppose that the normalized wavefunction for an electron in a carbon nanotube of length $L=10.0 \mathrm{~nm}$ is: $\psi=(2 / L)^{1 / 2} \sin (\pi x / L)$. Calculate the probability that the electron is between $x=4.95 \mathrm{~nm}$ and $5.05 \mathrm{~nm}$. | 0.020 | problem | 0.020 | 5.1(a) | matter |
||
A sample of the sugar D-ribose of mass $0.727 \mathrm{~g}$ was placed in a calorimeter and then ignited in the presence of excess oxygen. The temperature rose by $0.910 \mathrm{~K}$. In a separate experiment in the same calorimeter, the combustion of $0.825 \mathrm{~g}$ of benzoic acid, for which the internal energy of combustion is $-3251 \mathrm{~kJ} \mathrm{~mol}^{-1}$, gave a temperature rise of $1.940 \mathrm{~K}$. Calculate the enthalpy of formation of D-ribose. | -1270 | problem | -1270 | 57.1 | $\mathrm{~kJ} \mathrm{~mol}^{-1}$ | matter |
|
An electron confined to a metallic nanoparticle is modelled as a particle in a one-dimensional box of length $L$. If the electron is in the state $n=1$, calculate the probability of finding it in the following regions: $0 \leq x \leq \frac{1}{2} L$. | $\frac{1}{2}$ | problem | 0.5 | 9.3(a) | matter |
||
The carbon-carbon bond length in diamond is $154.45 \mathrm{pm}$. If diamond were considered to be a close-packed structure of hard spheres with radii equal to half the bond length, what would be its expected density? The diamond lattice is face-centred cubic and its actual density is $3.516 \mathrm{~g} \mathrm{~cm}^{-3}$. | 7.654 | problem | 7.654 | 38.3 | $\mathrm{~g} \mathrm{~cm}^{-3}$ | matter |
|
A swimmer enters a gloomier world (in one sense) on diving to greater depths. Given that the mean molar absorption coefficient of seawater in the visible region is $6.2 \times 10^{-3} \mathrm{dm}^3 \mathrm{~mol}^{-1} \mathrm{~cm}^{-1}$, calculate the depth at which a diver will experience half the surface intensity of light. | 0.87 | 0.87 | 40.7(a) | $\mathrm{~m}$ | matter |
||
Calculate the molar energy required to reverse the direction of an $\mathrm{H}_2 \mathrm{O}$ molecule located $100 \mathrm{pm}$ from a $\mathrm{Li}^{+}$ ion. Take the magnitude of the dipole moment of water as $1.85 \mathrm{D}$. | 1.07 | 1.07 | 35.1(a) | $10^3 \mathrm{~kJ} \mathrm{~mol}^{-1}$ | matter |
||
In an industrial process, nitrogen is heated to $500 \mathrm{~K}$ at a constant volume of $1.000 \mathrm{~m}^3$. The gas enters the container at $300 \mathrm{~K}$ and $100 \mathrm{~atm}$. The mass of the gas is $92.4 \mathrm{~kg}$. Use the van der Waals equation to determine the approximate pressure of the gas at its working temperature of $500 \mathrm{~K}$. For nitrogen, $a=1.39 \mathrm{dm}^6 \mathrm{~atm} \mathrm{~mol}^{-2}, b=0.0391 \mathrm{dm}^3 \mathrm{~mol}^{-1}$. | 140 | 140 | 36.6(a) | $\mathrm{~atm}$ | matter |
||
The chemical shift of the $\mathrm{CH}_3$ protons in acetaldehyde (ethanal) is $\delta=2.20$ and that of the $\mathrm{CHO}$ proton is 9.80 . What is the difference in local magnetic field between the two regions of the molecule when the applied field is $1.5 \mathrm{~T}$ | 11 | 11 | 48.2(a) | $\mu \mathrm{T}$ | matter |
||
Suppose that the junction between two semiconductors can be represented by a barrier of height $2.0 \mathrm{eV}$ and length $100 \mathrm{pm}$. Calculate the transmission probability of an electron with energy $1.5 \mathrm{eV}$. | 0.8 | 0.8 | 10.1(a) | matter |
|||
The diffusion coefficient of a particular kind of t-RNA molecule is $D=1.0 \times 10^{-11} \mathrm{~m}^2 \mathrm{~s}^{-1}$ in the medium of a cell interior. How long does it take molecules produced in the cell nucleus to reach the walls of the cell at a distance $1.0 \mu \mathrm{m}$, corresponding to the radius of the cell? | 1.7 | problem | 1.7 | 81.11 | $10^{-2} \mathrm{~s}$ | matter |
|
At what pressure does the mean free path of argon at $20^{\circ} \mathrm{C}$ become comparable to the diameter of a $100 \mathrm{~cm}^3$ vessel that contains it? Take $\sigma=0.36 \mathrm{~nm}^2$ | 0.195 | 0.195 | 78.6(a) | $\mathrm{Pa}$ | matter |
||
The equilibrium pressure of $\mathrm{O}_2$ over solid silver and silver oxide, $\mathrm{Ag}_2 \mathrm{O}$, at $298 \mathrm{~K}$ is $11.85 \mathrm{~Pa}$. Calculate the standard Gibbs energy of formation of $\mathrm{Ag}_2 \mathrm{O}(\mathrm{s})$ at $298 \mathrm{~K}$. | -11.2 | -11.2 | 73.4(a) | $\mathrm{~kJ} \mathrm{~mol}^{-1}$ | matter |
||
When alkali metals dissolve in liquid ammonia, their atoms each lose an electron and give rise to a deep-blue solution that contains unpaired electrons occupying cavities in the solvent. These 'metal-ammonia solutions' have a maximum absorption at $1500 \mathrm{~nm}$. Supposing that the absorption is due to the excitation of an electron in a spherical square well from its ground state to the next-higher state (see the preceding problem for information), what is the radius of the cavity? | 0.69 | problem | 0.69 | 11.5 | $\mathrm{~nm}$ | matter |
|
Electron diffraction makes use of electrons with wavelengths comparable to bond lengths. To what speed must an electron be accelerated for it to have a wavelength of $100 \mathrm{pm}$ ? | 7.27 | 7.27 | 4.8(a) | $10^6 \mathrm{~m} \mathrm{~s}^{-1}$ | matter |
||
Nelson, et al. (Science 238, 1670 (1987)) examined several weakly bound gas-phase complexes of ammonia in search of examples in which the $\mathrm{H}$ atoms in $\mathrm{NH}_3$ formed hydrogen bonds, but found none. For example, they found that the complex of $\mathrm{NH}_3$ and $\mathrm{CO}_2$ has the carbon atom nearest the nitrogen (299 pm away): the $\mathrm{CO}_2$ molecule is at right angles to the $\mathrm{C}-\mathrm{N}$ 'bond', and the $\mathrm{H}$ atoms of $\mathrm{NH}_3$ are pointing away from the $\mathrm{CO}_2$. The magnitude of the permanent dipole moment of this complex is reported as $1.77 \mathrm{D}$. If the $\mathrm{N}$ and $\mathrm{C}$ atoms are the centres of the negative and positive charge distributions, respectively, what is the magnitude of those partial charges (as multiples of $e$ )? | 0.123 | problem | 0.123 | 34.5 | matter |
||
The NOF molecule is an asymmetric rotor with rotational constants $3.1752 \mathrm{~cm}^{-1}, 0.3951 \mathrm{~cm}^{-1}$, and $0.3505 \mathrm{~cm}^{-1}$. Calculate the rotational partition function of the molecule at $25^{\circ} \mathrm{C}$. | 7.97 | 7.97 | 52.4(a) | $10^3$ | matter |
||
Suppose that $2.5 \mathrm{mmol} \mathrm{N}_2$ (g) occupies $42 \mathrm{~cm}^3$ at $300 \mathrm{~K}$ and expands isothermally to $600 \mathrm{~cm}^3$. Calculate $\Delta G$ for the process. | -17 | -17 | 66.1(a) | $\mathrm{~J}$ | matter |
||
Calculate the standard potential of the $\mathrm{Ce}^{4+} / \mathrm{Ce}$ couple from the values for the $\mathrm{Ce}^{3+} / \mathrm{Ce}$ and $\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}$ couples.
| -1.46 | -1.46 | 77.1(a) | $\mathrm{V}$ | matter |
||
An effusion cell has a circular hole of diameter $1.50 \mathrm{~mm}$. If the molar mass of the solid in the cell is $300 \mathrm{~g} \mathrm{~mol}^{-1}$ and its vapour pressure is $0.735 \mathrm{~Pa}$ at $500 \mathrm{~K}$, by how much will the mass of the solid decrease in a period of $1.00 \mathrm{~h}$ ? | 16 | 16 | 78.11(a) | $\mathrm{mg}$ | matter |
||
The speed of a certain proton is $6.1 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$. If the uncertainty in its momentum is to be reduced to 0.0100 per cent, what uncertainty in its location must be tolerated? | 52 | 52 | 8.1(a) | $\mathrm{pm}$ | matter |
||
It is possible to produce very high magnetic fields over small volumes by special techniques. What would be the resonance frequency of an electron spin in an organic radical in a field of $1.0 \mathrm{kT}$ ? | 2.8 | problem | 2.8 | 50.1 | $10^{13} \mathrm{~Hz}$ | matter |
|
A particle of mass $1.0 \mathrm{~g}$ is released near the surface of the Earth, where the acceleration of free fall is $g=8.91 \mathrm{~m} \mathrm{~s}^{-2}$. What will be its kinetic energy after $1.0 \mathrm{~s}$. Ignore air resistance? | 48 | 48 | 2.1(a) | $\mathrm{~mJ}$ | matter |
||
The flux of visible photons reaching Earth from the North Star is about $4 \times 10^3 \mathrm{~mm}^{-2} \mathrm{~s}^{-1}$. Of these photons, 30 per cent are absorbed or scattered by the atmosphere and 25 per cent of the surviving photons are scattered by the surface of the cornea of the eye. A further 9 per cent are absorbed inside the cornea. The area of the pupil at night is about $40 \mathrm{~mm}^2$ and the response time of the eye is about $0.1 \mathrm{~s}$. Of the photons passing through the pupil, about 43 per cent are absorbed in the ocular medium. How many photons from the North Star are focused onto the retina in $0.1 \mathrm{~s}$ ? | 4.4 | problem | 4.4 | 40.3 | $10^3$ | matter |
|
When ultraviolet radiation of wavelength $58.4 \mathrm{~nm}$ from a helium lamp is directed on to a sample of krypton, electrons are ejected with a speed of $1.59 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$. Calculate the ionization energy of krypton.
| 14 | 14 | 17.2(a) | $\mathrm{eV}$ | matter |
||
If $125 \mathrm{~cm}^3$ of hydrogen gas effuses through a small hole in 135 seconds, how long will it take the same volume of oxygen gas to effuse under the same temperature and pressure? | 537 | 537 | 78.10(a) | $\mathrm{s}$ | matter |
||
The vibrational wavenumber of $\mathrm{Br}_2$ is $323.2 \mathrm{~cm}^{-1}$. By evaluating the vibrational partition function explicitly (without approximation), at what temperature is the value within 5 per cent of the value calculated from the approximate formula? | 4500 | 4500 | 52.10(a) | $\mathrm{~K}$ | matter |
||
A thermodynamic study of $\mathrm{DyCl}_3$ (E.H.P. Cordfunke, et al., J. Chem. Thermodynamics 28, 1387 (1996)) determined its standard enthalpy of formation from the following information
(1) $\mathrm{DyCl}_3(\mathrm{~s}) \rightarrow \mathrm{DyCl}_3(\mathrm{aq}$, in $4.0 \mathrm{M} \mathrm{HCl}) \quad \Delta_{\mathrm{r}} H^{\ominus}=-180.06 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(2) $\mathrm{Dy}(\mathrm{s})+3 \mathrm{HCl}(\mathrm{aq}, 4.0 \mathrm{~m}) \rightarrow \mathrm{DyCl}_3(\mathrm{aq}$, in $4.0 \mathrm{M} \mathrm{HCl}(\mathrm{aq}))+\frac{3}{2} \mathrm{H}_2(\mathrm{~g})$ $\Delta_{\mathrm{r}} H^{\ominus}=-699.43 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(3) $\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{aq}, 4.0 \mathrm{M}) \quad \Delta_{\mathrm{r}} H^{\ominus}=-158.31 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Determine $\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{DyCl}_3, \mathrm{~s}\right)$ from these data. | -994.3 | problem | -994.3 | 57.5 | $\mathrm{~kJ} \mathrm{~mol}^{-1}$ | matter |
|
Calculate $\Delta_{\mathrm{r}} G^{\ominus}(375 \mathrm{~K})$ for the reaction $2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})$ from the values of $\Delta_{\mathrm{r}} G^{\ominus}(298 \mathrm{~K})$ : and $\Delta_{\mathrm{r}} H^{\ominus}(298 \mathrm{~K})$, and the GibbsHelmholtz equation. | -501 | problem | -501 | 66.1 | $\mathrm{~kJ} \mathrm{~mol}^{-1}$ | matter |
|
The vapour pressure of benzene is $53.3 \mathrm{kPa}$ at $60.6^{\circ} \mathrm{C}$, but it fell to $51.5 \mathrm{kPa}$ when $19.0 \mathrm{~g}$ of an non-volatile organic compound was dissolved in $500 \mathrm{~g}$ of benzene. Calculate the molar mass of the compound. | 85 | 85 | 70.8(a) | $\mathrm{~g} \mathrm{~mol}^{-1}$ | matter |
||
J.G. Dojahn, et al. (J. Phys. Chem. 100, 9649 (1996)) characterized the potential energy curves of the ground and electronic states of homonuclear diatomic halogen anions. The ground state of $\mathrm{F}_2^{-}$ is ${ }^2 \sum_{\mathrm{u}}^{+}$ with a fundamental vibrational wavenumber of $450.0 \mathrm{~cm}^{-1}$ and equilibrium internuclear distance of $190.0 \mathrm{pm}$. The first two excited states are at 1.609 and $1.702 \mathrm{eV}$ above the ground state. Compute the standard molar entropy of $\mathrm{F}_2^{-}$ at $ 298 \mathrm{~K}$. | 199.4 | problem | 199.4 | 60.3 | $\mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ | matter |
|
The duration of a $90^{\circ}$ or $180^{\circ}$ pulse depends on the strength of the $\mathscr{B}_1$ field. If a $180^{\circ}$ pulse requires $12.5 \mu \mathrm{s}$, what is the strength of the $\mathscr{B}_1$ field? | 5.9 | 5.9 | 49.1(a) | $10^{-4} \mathrm{~T}$ | matter |
||
In 1976 it was mistakenly believed that the first of the 'superheavy' elements had been discovered in a sample of mica. Its atomic number was believed to be 126. What is the most probable distance of the innermost electrons from the nucleus of an atom of this element? (In such elements, relativistic effects are very important, but ignore them here.) | 0.42 | problem | 0.42 | 18.1 | $\mathrm{pm}$ | matter |
|
The ground level of $\mathrm{Cl}$ is ${ }^2 \mathrm{P}_{3 / 2}$ and a ${ }^2 \mathrm{P}_{1 / 2}$ level lies $881 \mathrm{~cm}^{-1}$ above it. Calculate the electronic contribution to the molar Gibbs energy of $\mathrm{Cl}$ atoms at $500 \mathrm{~K}$. | -6.42 | -6.42 | 64.5(a) | $\mathrm{~kJ} \mathrm{~mol}^{-1}$ | matter |
||
Calculate the melting point of ice under a pressure of 50 bar. Assume that the density of ice under these conditions is approximately $0.92 \mathrm{~g} \mathrm{~cm}^{-3}$ and that of liquid water is $1.00 \mathrm{~g} \mathrm{~cm}^{-3}$. | 272.8 | 272.8 | 69.9(a) | $\mathrm{K}$ | matter |
||
What is the temperature of a two-level system of energy separation equivalent to $400 \mathrm{~cm}^{-1}$ when the population of the upper state is one-third that of the lower state? | 524 | 524 | 51.4(a) | $ \mathrm{~K}$ | matter |
||
At $300 \mathrm{~K}$ and $20 \mathrm{~atm}$, the compression factor of a gas is 0.86 . Calculate the volume occupied by $8.2 \mathrm{mmol}$ of the gas under these conditions. | 8.7 | problem | 8.7 | 36.3(a) | $\mathrm{~cm}^3$ | matter |
|
A very crude model of the buckminsterfullerene molecule $\left(\mathrm{C}_{60}\right)$ is to treat it as a collection of electrons in a cube with sides of length equal to the mean diameter of the molecule $(0.7 \mathrm{~nm})$. Suppose that only the $\pi$ electrons of the carbon atoms contribute, and predict the wavelength of the first excitation of $\mathrm{C}_{60}$. (The actual value is $730 \mathrm{~nm}$.) | 1.6 | problem | 1.6 | 11.3 | $\mu \mathrm{m}$ | matter |
|
From the expression for the work function $\Phi=h \nu-E_{\mathrm{k}}$ the minimum frequency for photoejection is
$$
\nu_{\min }=\frac{\Phi}{h}=\frac{h v-E_{\mathrm{k}}}{h} \stackrel{\nu=c \mid \lambda}{=} \frac{c}{\lambda}-\frac{E_{\mathrm{k}}}{h}
$$
The maximum wavelength is therefore
$$
\lambda_{\max }=\frac{c}{v_{\min }}=\frac{c}{c / \lambda-E_{\mathrm{k}} / h}=\frac{1}{1 / \lambda-E_{\mathrm{k}} / h c}
$$
Now we substitute the data. The kinetic energy of the electron is
$$
\begin{aligned}
& E_k=1.77 \mathrm{eV} \times\left(1.602 \times 10^{-19} \mathrm{JeV}^{-1}\right)=2.83 \ldots \times 10^{-19} \mathrm{~J} \\
& \frac{E_k}{h c}=\frac{2.83 \ldots \times 10^{-19} \mathrm{~J}}{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(2.998 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}=1.42 \ldots \times 10^6 \mathrm{~m}^{-1}
\end{aligned}
$$
Therefore, with
$$
\begin{aligned}
& 1 / \lambda=1 / 305 \mathrm{~nm}=3.27 \ldots \times 10^6 \mathrm{~m}^{-1}, \\
& \lambda_{\max }=\frac{1}{\left(3.27 \ldots \times 10^6 \mathrm{~m}^{-1}\right)-\left(1.42 \ldots \times 10^6 \mathrm{~m}^{-1}\right)}=5.40 \times 10^{-7} \mathrm{~m}
\end{aligned}
$$
or $540 \mathrm{~nm}$.
| Calculating the maximum wavelength capable of photoejection
A photon of radiation of wavelength $305 \mathrm{~nm}$ ejects an electron from a metal with a kinetic energy of $1.77 \mathrm{eV}$. Calculate the maximum wavelength of radiation capable of ejecting an electron from the metal. | 540 | with solution | 540 | 4.1 | $\mathrm{~nm}$ | matter |
For van der Waals gas of $\mathrm{CO}_2$ , $a=3.610 \mathrm{dm}^6$ atm $\mathrm{mol}^{-2}$ and $b=4.29 \times 10^{-2} \mathrm{dm}^3 \mathrm{~mol}^{-1}$. Under the stated conditions, $R T / p=0.410 \mathrm{dm}^3 \mathrm{~mol}^{-1}$. The coefficients in the equation for $V_{\mathrm{m}}$ are therefore
$$
\begin{aligned}
b+R T / p & =0.453 \mathrm{dm}^3 \mathrm{~mol}^{-1} \\
a / p & =3.61 \times 10^{-2}\left(\mathrm{dm}^3 \mathrm{~mol}^{-1}\right)^2 \\
a b / p & =1.55 \times 10^{-3}\left(\mathrm{dm}^3 \mathrm{~mol}^{-1}\right)^3
\end{aligned}
$$
Therefore, on writing $x=V_{\mathrm{m}} /\left(\mathrm{dm}^3 \mathrm{~mol}^{-1}\right)$, the equation to solve is
$$
x^3-0.453 x^2+\left(3.61 \times 10^{-2}\right) x-\left(1.55 \times 10^{-3}\right)=0
$$
The acceptable root is $x=0.366$, which implies that $V_{\mathrm{m}}=0.366$ $\mathrm{dm}^3 \mathrm{~mol}^{-1}$. The molar volume of a perfect gas under these conditions is $0.410 \mathrm{dm}^3 \mathrm{~mol}^{-1}$.
| Estimate the molar volume of $\mathrm{CO}_2$ at $500 \mathrm{~K}$ and 100 atm by treating it as a van der Waals gas. | 0.366 | with solution | 0.366 | 36.1 | $\mathrm{dm}^3 \mathrm{~mol}^{-1}$ | matter |
Replacing $\mu$ by $m_{\mathrm{e}}$ and using $\hbar=h / 2 \pi$, we can write the expression for the energy as
$$
E_n=-\frac{Z^2 m_e e^4}{8 \varepsilon_0^2 h^2 n^2}=-\frac{Z^2 h c \tilde{R}_{\infty}}{n^2}
$$
with
$$
\tilde{R}_{\infty}=\frac{9.10938 \times 10^{-31} \mathrm{~kg} \times (1.602176 \times 10^{-19} \mathrm{C})^4}{8 \times (8.85419 \times 10^{-12} \mathrm{~J}^{-1} \mathrm{C}^2 \mathrm{~m}^{-1} )^2 \times(6.62608 \times 10^{-34} \mathrm{Js})^3} \times \underbrace{2.997926 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}}_c =109737 \mathrm{~cm}^{-1} $$
and
$$
\begin{aligned}
h c \tilde{R}_{\infty}= & \left(6.62608 \times 10^{-34} \mathrm{Js}\right) \times\left(2.997926 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right) \\
& \times\left(109737 \mathrm{~cm}^{-1}\right) \\
= & 2.17987 \times 10^{-18} \mathrm{~J}
\end{aligned}
$$
Therefore, for $n=3$, the energy is
$$
\begin{aligned}
& E_3=-\frac{\overbrace{4}^{Z^2} \times \overbrace{2.17987 \times 10^{-18} \mathrm{~J}}^{h c \tilde{R}_{\infty}}}{\underset{\tilde{n}^2}{9}} \\
& =-9.68831 \times 10^{-19} \mathrm{~J} \\
&
\end{aligned}
$$
or $-0.968831 \mathrm{aJ}$ (a, for atto, is the prefix that denotes $10^{-18}$ ). In some applications it is useful to express the energy in electronvolts $\left(1 \mathrm{eV}=1.602176 \times 10^{-19} \mathrm{~J}\right)$; in this case, $E_3=-6.04697 \mathrm{eV}$
| The single electron in a certain excited state of a hydrogenic $\mathrm{He}^{+}$ion $(Z=2)$ is described by the wavefunction $R_{3,2}(r) \times$ $Y_{2,-1}(\theta, \phi)$. What is the energy of its electron? | -6.04697 | with solution | -6.04697 | 17.1 | $ \mathrm{eV}$ | matter |
From the equipartition principle, we know that the mean translational kinetic energy of a neutron at a temperature $T$ travelling in the $x$-direction is $E_{\mathrm{k}}=\frac{1}{2} k T$. The kinetic energy is also equal to $p^2 / 2 m$, where $p$ is the momentum of the neutron and $m$ is its mass. Hence, $p=(m k T)^{1 / 2}$. It follows from the de Broglie relation $\lambda=h / p$ that the neutron's wavelength is
$$
\lambda=\frac{h}{(m k T)^{1 / 2}}
$$
Therefore, at $373 \mathrm{~K}$,
$$
\begin{aligned}
\lambda & =\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left\{\left(1.675 \times 10^{-27} \mathrm{~kg}\right) \times\left(1.381 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right) \times(373 \mathrm{~K})\right\}^{1 / 2}} \\
& =\frac{6.626 \times 10^{-34}}{\left(1.675 \times 10^{-27} \times 1.381 \times 10^{-23} \times 373\right)^{1 / 2}} \frac{\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-1}}{\left(\mathrm{~kg}^2 \mathrm{~m}^2 \mathrm{~s}^{-2}\right)^{1 / 2}} \\
& =2.26 \times 10^{-10} \mathrm{~m}=226 \mathrm{pm}
\end{aligned}
$$
where we have used $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}$. | Calculate the typical wavelength of neutrons after reaching thermal equilibrium with their surroundings at $373 \mathrm{~K}$. For simplicity, assume that the particles are travelling in one dimension. | 226 | with solution | 226 | 37.4 | $\mathrm{pm}$ | matter |
The amount of $\mathrm{N}_2$ molecules (of molar mass $28.02 \mathrm{~g}$ $\mathrm{mol}^{-1}$ ) present is
$$
n\left(\mathrm{~N}_2\right)=\frac{m}{M\left(\mathrm{~N}_2\right)}=\frac{1.25 \mathrm{~g}}{28.02 \mathrm{~g} \mathrm{~mol}^{-1}}=\frac{1.25}{28.02} \mathrm{~mol}
$$
The temperature of the sample is
$$
T / K=20+273.15, \text { so } T=(20+273.15) \mathrm{K}
$$
Therefore, after rewriting eqn 1.5 as $p=n R T / V$,
$$
\begin{aligned}
p & =\frac{\overbrace{(1.25 / 28.02) \mathrm{mol}}^n \times \overbrace{\left(8.3145 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)}^R \times \overbrace{(20+273.15) \mathrm{K}}^{\left(2.50 \times 10^{-4}\right) \mathrm{m}^3}}{\underbrace{R}_V} \\
& =\frac{(1.25 / 28.02) \times(8.3145) \times(20+273.15)}{2.50 \times 10^{-4}} \frac{\mathrm{J}}{\mathrm{m}^3} \\
1 \mathrm{Jm}^{-3} & =1 \mathrm{~Pa} \\
& \stackrel{\infty}{=} 4.35 \times 10^5 \mathrm{~Pa}=435 \mathrm{kPa}
\end{aligned}
$$
| Using the perfect gas equation
Calculate the pressure in kilopascals exerted by $1.25 \mathrm{~g}$ of nitrogen gas in a flask of volume $250 \mathrm{~cm}^3$ at $20^{\circ} \mathrm{C}$. | 435 | with equation | 435 | 1.1 | $\mathrm{kPa}$ | matter |
First, note that
$$
\frac{\hbar^2}{2 I}=\frac{\left(1.055 \times 10^{-34} \mathrm{Js}^2\right.}{2 \times\left(2.6422 \times 10^{-47} \mathrm{~kg} \mathrm{~m}^2\right)}=2.106 \ldots \times 10^{-22} \mathrm{~J}
$$
or $0.2106 \ldots$ zJ. We now draw up the following table, where the molar energies are obtained by multiplying the individual energies by Avogadro's constant:
\begin{tabular}{llll}
\hline$J$ & $E / z J$ & $E /\left(\mathrm{J} \mathrm{mol}^{-1}\right)$ & Degeneracy \\
\hline 0 & 0 & 0 & 1 \\
1 & 0.4212 & 253.6 & 3 \\
2 & 1.264 & 760.9 & 5 \\
3 & 2.527 & 1522 & 7 \\
\hline
\end{tabular}
The energy separation between the two lowest rotational energy levels $\left(J=0\right.$ and 1 ) is $4.212 \times 10^{-22} \mathrm{~J}$, which corresponds to a photon frequency of
$$
\nu=\frac{\Delta E}{h}=\frac{4.212 \times 10^{-22} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}=6.357 \times 10^{11} \mathrm{~s}^{-1}=635.7 \mathrm{GHz}
$$ | Determine the energies and degeneracies of the lowest four energy levels of an ${ }^1 \mathrm{H}^{35} \mathrm{Cl}$ molecule freely rotating in three dimensions. What is the frequency of the transition between the lowest two rotational levels? The moment of inertia of an ${ }^1 \mathrm{H}^{35} \mathrm{Cl}$ molecule is $2.6422 \times 10^{-47} \mathrm{~kg} \mathrm{~m}^2$.
| 635.7 | with solution | 635.7 | 14.1 | $\mathrm{GHz}$ | matter |
At a temperature $T$, the ratio of the spectral density of states at a wavelength $\lambda_1$ to that at $\lambda_2$ is given by
$$
\frac{\rho\left(\lambda_1, T\right)}{\rho\left(\lambda_2, T\right)}=\left(\frac{\lambda_2}{\lambda_1}\right)^5 \times \frac{\left(\mathrm{e}^{h c / \lambda_2 k T}-1\right)}{\left(\mathrm{e}^{h c / \lambda_1 k T}-1\right)}
$$
Insert the data and evaluate this ratio.
Answer With $\lambda_1=450 \mathrm{~nm}$ and $\lambda_2=700 \mathrm{~nm}$,
$$
\begin{aligned}
\frac{h c}{\lambda_1 k T} & =\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(2.998 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(450 \times 10^{-9} \mathrm{~m}\right) \times\left(1.381 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right) \times(298 \mathrm{~K})}=107.2 \ldots \\
\frac{h c}{\lambda_2 k T} & =\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(2.998 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(700 \times 10^{-9} \mathrm{~m}\right) \times\left(1.381 \times 10^{-23} \mathrm{JK}^{-1}\right) \times(298 \mathrm{~K})}=68.9 \ldots
\end{aligned}
$$
and therefore
$$
\begin{aligned}
& \frac{\rho(450 \mathrm{~nm}, 298 \mathrm{~K})}{\rho(700 \mathrm{~nm}, 298 \mathrm{~K})}=\left(\frac{700 \times 10^{-9} \mathrm{~m}}{450 \times 10^{-9} \mathrm{~m}}\right)^5 \times \frac{\left(\mathrm{e}^{68.9 \cdots}-1\right)}{\left(\mathrm{e}^{107.2 \cdots}-1\right)} \\
& =9.11 \times\left(2.30 \times 10^{-17}\right)=2.10 \times 10^{-16}
\end{aligned}
$$ | Using the Planck distribution
Compare the energy output of a black-body radiator (such as an incandescent lamp) at two different wavelengths by calculating the ratio of the energy output at $450 \mathrm{~nm}$ (blue light) to that at $700 \mathrm{~nm}$ (red light) at $298 \mathrm{~K}$.
| 2.10 | with solution | 2.10 | 4.1 | $10^{-16}$ | matter |
Rearrangement of eqn $$
\mathcal{H}_{\mathrm{c}}(T)=\mathcal{H}_{\mathrm{c}}(0) (1-\frac{T^2}{T_{\mathrm{c}}^2})
$$ gives
$$
T=T_{\mathrm{c}}\left(1-\frac{\mathcal{H}_{\mathrm{c}}(T)}{\mathcal{H}_{\mathrm{c}}(0)}\right)^{1 / 2}
$$
and substitution of the data gives
$$
T=(7.19 \mathrm{~K}) \times\left(1-\frac{20 \mathrm{kAm}^{-1}}{63.9 \mathrm{kAm}^{-1}}\right)^{1 / 2}=6.0 \mathrm{~K}
$$
That is, lead becomes superconducting at temperatures below $6.0 \mathrm{~K}$.
| Lead has $T_{\mathrm{c}}=7.19 \mathrm{~K}$ and $\mathcal{H}_{\mathrm{c}}(0)=63.9 \mathrm{kA} \mathrm{m}^{-1}$. At what temperature does lead become superconducting in a magnetic field of $20 \mathrm{kA} \mathrm{m}^{-1}$ ? | 6.0 | with solution | 6.0 | 39.2 | $\mathrm{~K}$ | matter |
The wavenumber of the photon emitted when an electron makes a transition from $n_2=2$ to $n_1=1$ is given by
$$
\begin{aligned}
\tilde{\boldsymbol{v}} & =-\tilde{R}_{\mathrm{H}}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\
& =-\left(109677 \mathrm{~cm}^{-1}\right) \times\left(\frac{1}{2^2}-\frac{1}{1^2}\right) \\
& =82258 \mathrm{~cm}^{-1}
\end{aligned}
$$ | When an electric discharge is passed through gaseous hydrogen, the $\mathrm{H}_2$ molecules are dissociated and energetically excited $\mathrm{H}$ atoms are produced. If the electron in an excited $\mathrm{H}$ atom makes a transition from $n=2$ to $n=1$, calculate the wavenumber of the corresponding line in the emission spectrum. | 82258 | with solution | 82258 | 17.2 | $\mathrm{~cm}^{-1}$ | matter |
The wavefunction for a hydrogen 1 s orbital is
$$
\psi=\left(\frac{1}{\pi a_0^3}\right)^{1 / 2} \mathrm{e}^{-r / a_0}
$$
so, because $\mathrm{d} \tau=r^2 \mathrm{~d} r \sin \theta \mathrm{d} \theta \mathrm{d} \phi$, the expectation value of $1 / r$ is written as
$$
\begin{aligned}
\left\langle\frac{1}{r}\right\rangle & =\int \frac{\psi^* \psi}{r} \mathrm{~d} \tau=\frac{1}{\pi a_0^3} \int_0^{2 \pi} \mathrm{d} \phi \int_0^\pi \sin \theta \mathrm{d} \theta \int_0^{\infty} r \mathrm{e}^{-2 r / a_0} \mathrm{~d} r \\
& =\frac{4}{a_0^3} \overbrace{\int_0^{\infty} r \mathrm{e}^{-2 r / a_0} \mathrm{~d} r}^{a_0^2 / 4 \text { (Integral E.1) }}=\frac{1}{a_0}
\end{aligned}
$$
where we used the integral listed in the Resource section. Therefore,
$$
\begin{aligned}
& =\frac{\left(1.602 \times 10^{-19} \mathrm{C}\right)^2 \times(4 \pi \times 10^{-7} \overbrace{\mathrm{J}}^{\mathrm{Jg} \mathrm{m}^2 \mathrm{~s}^{-2}} \mathrm{~s}^2 \mathrm{C}^{-2} \mathrm{~m}^{-1})}{12 \pi \times\left(9.109 \times 10^{-31} \mathrm{~kg}\right) \times\left(5.292 \times 10^{-11} \mathrm{~m}\right)} \\
& =1.775 \times 10^{-5} \\
&
\end{aligned}
$$
| Calculate the shielding constant for the proton in a free $\mathrm{H}$ atom. | 1.775 | with solution | 1.775 | 48.2 | $10^{-5}$ | matter |
Use the $D_0$ value of $\mathrm{H}_2(4.478 \mathrm{eV})$ and the $D_0$ value of $\mathrm{H}_2^{+}(2.651 \mathrm{eV})$ to calculate the first ionization energy of $\mathrm{H}_2$ (that is, the energy needed to remove an electron from $\mathrm{H}_2$ ). | 15.425 | 15.425 | 13.3 | $\mathrm{eV}$ | quan |
||
Calculate the energy of one mole of UV photons of wavelength $300 \mathrm{~nm}$ and compare it with a typical single-bond energy of $400 \mathrm{~kJ} / \mathrm{mol}$. | 399 | 399 | 1.3 | $\mathrm{~kJ} / \mathrm{mol}$ | quan |
||
Calculate the magnitude of the spin angular momentum of a proton. Give a numerical answer. | 9.13 | 9.13 | 10.1 | $10^{-35} \mathrm{~J} \mathrm{~s}$ | quan |
||
The ${ }^7 \mathrm{Li}^1 \mathrm{H}$ ground electronic state has $D_0=2.4287 \mathrm{eV}, \nu_e / c=1405.65 \mathrm{~cm}^{-1}$, and $\nu_e x_e / c=23.20 \mathrm{~cm}^{-1}$, where $c$ is the speed of light. (These last two quantities are usually designated $\omega_e$ and $\omega_e x_e$ in the literature.) Calculate $D_e$ for ${ }^7 \mathrm{Li}^1 \mathrm{H}$. | 2.5151 | 2.5151 | 13.5 | $\mathrm{eV}$ | quan |
||
The positron has charge $+e$ and mass equal to the electron mass. Calculate in electronvolts the ground-state energy of positronium-an "atom" that consists of a positron and an electron. | -6.8 | -6.8 | 6.22 | $\mathrm{eV}$ | quan |
||
What is the value of the angular-momentum quantum number $l$ for a $t$ orbital? | 14 | 14 | 6.29 | quan |
|||
How many states belong to the carbon configurations $1 s^2 2 s^2 2 p^2$? | 15 | 15 | 11.22 | quan |
|||
Calculate the energy needed to compress three carbon-carbon single bonds and stretch three carbon-carbon double bonds to the benzene bond length $1.397 Å$. Assume a harmonicoscillator potential-energy function for bond stretching and compression. Typical carboncarbon single- and double-bond lengths are 1.53 and $1.335 Å$; typical stretching force constants for carbon-carbon single and double bonds are 500 and $950 \mathrm{~N} / \mathrm{m}$. | 27 | Angstrom | 27 | 17.9 | $\mathrm{kcal} / \mathrm{mol}$ | quan |
|
When a particle of mass $9.1 \times 10^{-28} \mathrm{~g}$ in a certain one-dimensional box goes from the $n=5$ level to the $n=2$ level, it emits a photon of frequency $6.0 \times 10^{14} \mathrm{~s}^{-1}$. Find the length of the box. | 1.8 | 1.8 | 2.13 | $\mathrm{~nm}$ | quan |
||
Use the normalized Numerov-method harmonic-oscillator wave functions found by going from -5 to 5 in steps of 0.1 to estimate the probability of being in the classically forbidden region for the $v=0$ state. | 0.16 | 0.16 | 4.42 | quan |
|||
Calculate the de Broglie wavelength of an electron moving at 1/137th the speed of light. (At this speed, the relativistic correction to the mass is negligible.) | 0.332 | 0.332 | 1.6 | $\mathrm{~nm}$ | quan |
||
Calculate the angle that the spin vector $S$ makes with the $z$ axis for an electron with spin function $\alpha$. | 54.7 | 54.7 | 10.2 | $^{\circ}$ | quan |
||
The AM1 valence electronic energies of the atoms $\mathrm{H}$ and $\mathrm{O}$ are $-11.396 \mathrm{eV}$ and $-316.100 \mathrm{eV}$, respectively. For $\mathrm{H}_2 \mathrm{O}$ at its AM1-calculated equilibrium geometry, the AM1 valence electronic energy (core-core repulsion omitted) is $-493.358 \mathrm{eV}$ and the AM1 core-core repulsion energy is $144.796 \mathrm{eV}$. For $\mathrm{H}(g)$ and $\mathrm{O}(g), \Delta H_{f, 298}^{\circ}$ values are 52.102 and $59.559 \mathrm{kcal} / \mathrm{mol}$, respectively. Find the AM1 prediction of $\Delta H_{f, 298}^{\circ}$ of $\mathrm{H}_2 \mathrm{O}(g)$. | -59.24 | -59.24 | 17.29 | $\mathrm{kcal} / \mathrm{mol}$ | quan |
||
Given that $D_e=4.75 \mathrm{eV}$ and $R_e=0.741 Å$ for the ground electronic state of $\mathrm{H}_2$, find $U\left(R_e\right)$ for this state. | -31.95 | Angstrom | -31.95 | 14.35 | $\mathrm{eV}$ | quan |
|
For $\mathrm{NaCl}, R_e=2.36 Å$. The ionization energy of $\mathrm{Na}$ is $5.14 \mathrm{eV}$, and the electron affinity of $\mathrm{Cl}$ is $3.61 \mathrm{eV}$. Use the simple model of $\mathrm{NaCl}$ as a pair of spherical ions in contact to estimate $D_e$. [One debye (D) is $3.33564 \times 10^{-30} \mathrm{C} \mathrm{m}$.] | 4.56 | Angstrom | 4.56 | 14.5 | $\mathrm{eV}$ | quan |
|
Find the number of CSFs in a full CI calculation of $\mathrm{CH}_2 \mathrm{SiHF}$ using a 6-31G** basis set. | 1.86 | 1.86 | 16.1 | $10^{28} $ | quan |
||
Calculate the ratio of the electrical and gravitational forces between a proton and an electron. | 2 | 2 | 6.15 | $10^{39}$ | quan |
||
A one-particle, one-dimensional system has the state function
$$
\Psi=(\sin a t)\left(2 / \pi c^2\right)^{1 / 4} e^{-x^2 / c^2}+(\cos a t)\left(32 / \pi c^6\right)^{1 / 4} x e^{-x^2 / c^2}
$$
where $a$ is a constant and $c=2.000 Å$. If the particle's position is measured at $t=0$, estimate the probability that the result will lie between $2.000 Å$ and $2.001 Å$. | 0.000216 | Angstrom | 0.000216 | 1.13 | quan |
||
The $J=2$ to 3 rotational transition in a certain diatomic molecule occurs at $126.4 \mathrm{GHz}$, where $1 \mathrm{GHz} \equiv 10^9 \mathrm{~Hz}$. Find the frequency of the $J=5$ to 6 absorption in this molecule. | 252.8 | Approximated answer | 252.8 | 6.10 | $\mathrm{GHz}$ | quan |
|
Assume that the charge of the proton is distributed uniformly throughout the volume of a sphere of radius $10^{-13} \mathrm{~cm}$. Use perturbation theory to estimate the shift in the ground-state hydrogen-atom energy due to the finite proton size. The potential energy experienced by the electron when it has penetrated the nucleus and is at distance $r$ from the nuclear center is $-e Q / 4 \pi \varepsilon_0 r$, where $Q$ is the amount of proton charge within the sphere of radius $r$. The evaluation of the integral is simplified by noting that the exponential factor in $\psi$ is essentially equal to 1 within the nucleus.
| 1.2 | 1.2 | 9.9 | $10^{-8} \mathrm{eV}$ | quan |
||
An electron in a three-dimensional rectangular box with dimensions of $5.00 Å, 3.00 Å$, and $6.00 Å$ makes a radiative transition from the lowest-lying excited state to the ground state. Calculate the frequency of the photon emitted. | 7.58 | Angstrom | 7.58 | 3.35 | $10^{14} \mathrm{~s}^{-1}$ | quan |
|
Do $\mathrm{HF} / 6-31 \mathrm{G}^*$ geometry optimizations on one conformers of $\mathrm{HCOOH}$ with $\mathrm{OCOH}$ dihedral angle of $0^{\circ}$. Calculate the dipole moment. | 1.41 | 1.41 | 15.57 | $\mathrm{D}$ | quan |
||
Frozen-core $\mathrm{SCF} / \mathrm{DZP}$ and CI-SD/DZP calculations on $\mathrm{H}_2 \mathrm{O}$ at its equilibrium geometry gave energies of -76.040542 and -76.243772 hartrees. Application of the Davidson correction brought the energy to -76.254549 hartrees. Find the coefficient of $\Phi_0$ in the normalized CI-SD wave function. | 0.9731 | 0.9731 | 16.3 | quan |
|||
Let $w$ be the variable defined as the number of heads that show when two coins are tossed simultaneously. Find $\langle w\rangle$. | 1 | 1 | 5.8 | quan |