Problem set 83

1. Suppose that the Coulomb potential of the hydrogen atom is changed by adding an inverse-square term such that the total potential is $V(\vec r)=-\frac{Ze^2}{r}+\frac{g}{r^2}$, where $g$ is a constant. The energy eigenvalues $E_{nlm}$ in the modified potential
1. depend on $n$ and $l$, but not on $m$
2. depend on $n$ but not on $l$ and $m$
3. depend on $n$ and $m$, but not on $l$
4. depend explicitly on all three quantum numbers $n$, $l$ and $m$

Radial equation for hydrogen atom is $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{\hbar^2l(l+1)}{2\mu r^2}R+V(r)R=ER}$$ For $V(\vec r)=-\frac{Ze^2}{r}+\frac{g}{r^2}$ equation becomes $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{\hbar^2l(l+1)}{2\mu r^2}R+V(r)R+\frac{g}{r^2}R=ER}$$ $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left[\frac{\hbar^2\left(l(l+1)+\frac{2\mu}{\hbar^2}g\right)}{2\mu r^2}\right]R+V(r)R=ER}$$ Let $l_{eff}(l_{eff}+1)=l(l+1)+\frac{2\mu}{\hbar^2}g$ $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left[\frac{\hbar^2l_{eff}(l_{eff}+1)}{2\mu r^2}\right]R+V(r)R=ER}$$ $$E\propto\frac{1}{(N+l_{eff}+1)^2}$$

Hence, answer is (A)

2. When an ideal monatomic gas is expanded adiabatically from an initial volume $V_0$ to $3V_0$, its temperature changes from $T_0$ to $T$. Then the ratio $T/T_0$is
1. $\frac{1}{3}$
2. $\left(\frac{1}{3}\right)^{2/3}$
3. $\left(\frac{1}{3}\right)^{1/3}$
4. 3

For adiabatic expansion of ideal gas we have $$PV^{\gamma}=const$$ $$TV^{\gamma-1}=const$$ $$P^{1-\gamma}T^{\gamma}=const$$ $$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$ $$\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$ For monoatomic ideal gas $\gamma=\frac{5}{3}$ $$\frac{T_2}{T_1}=\left(\frac{1}{3}\right)^{\frac{5}{3}-1}$$ $$\frac{T_2}{T_1}=\left(\frac{1}{3}\right)^{\frac{2}{3}}$$

Hence, answer is (B)

3. A box of volume $V$ containing $N$ molecules of an ideal gas, is divided by a wall with a hole into two compartments. If the volume of the smaller compartment is $V/3$, the variance of the number of particles in it, is
1. $N/3$
2. $2N/9$
3. $\sqrt{N}$
4. $\sqrt{N}/3$

When a box of $V$, containing $N$ molecules, is divided into two compartments of volume $pV$ and $qV$ such that $pV+qV=V$ (or $p+q=1$) then variance in number of particles is given by $$\sigma=Npq$$ In our case compartments have volume $\frac{1}{3}V$ and $\frac{2}{3}V$. Hence, $p=\frac{1}{3}$ and $q=\frac{2}{3}$ $$\sigma=N\frac{1}{3}\frac{2}{3}=\frac{2N}{9}$$

Hence, answer is (B)

4. A gas of non-relativistic classical particles in one dimension is subjected to a potential $V(x)=\alpha|x|$ (where $\alpha$ is a constant). The partition function is ($\beta=\frac{1}{k_BT}$)
1. $\sqrt{\frac{4m\pi}{\beta^3\alpha^2h^2}}$
2. $\sqrt{\frac{2m\pi}{\beta^3\alpha^2h^2}}$
3. $\sqrt{\frac{8m\pi}{\beta^3\alpha^2h^2}}$
4. $\sqrt{\frac{3m\pi}{\beta^3\alpha^2h^2}}$

Classical partition function for single particle is $$Z_1=\frac{1}{h^3}\int d^3p\:d^3q\:e^{-\beta H(q,p)}$$ For one dimension $$Z_1=\frac{1}{h}\int dp_x\:dx\:e^{-\beta H(q,p)}$$ $$H=\frac{p_x^2}{2m}+\alpha|x|$$ \begin{align*} Z_1&=\frac{1}{h}\int dp_x\:dx\:e^{-\beta \left(\frac{p_x^2}{2m}+\alpha|x|\right)}\\ &=\frac{1}{h}\left\{\int dp_x\:e^{-\frac{\beta}{2m} p_x^2}\!\!\int dx\:e^{-\beta \alpha|x|}\!\right\}\\ \end{align*} Using $\int dx\:e^{-a x^2}=\sqrt{\frac{\pi}{a}}$ \begin{align*} I_1&=\int dp_x\:e^{-\frac{\beta}{2m} p_x^2}\\ &=\sqrt{\frac{2m\pi}{\beta}} \end{align*} \begin{align*} I_2&=\int dx\:e^{-\beta \alpha|x|}\\ &=2\int_0^\infty dx\:e^{-\beta \alpha x}\\ &=2\frac{1}{\beta\alpha} \end{align*} $$Z_1=\frac{1}{h}\sqrt{\frac{2m\pi}{\beta}}\frac{2}{\beta\alpha}$$ $$Z_1=\sqrt{\frac{8m\pi}{\beta^3\alpha^2h^2}}$$

Hence, answer is (C)

5. The dependence of current $I$ on the voltage $V$ of a certain device is given by $$I=I_0\left(1-\frac{V}{V_0}\right)^2$$ where $I_0$ and $V_0$ are constants. In an experiment the current $I$ is measured as the voltage $V$ applied across the device is increased. The parameters $V_0$ and $\sqrt{I_0}$ can be graphically determined as
1. the slope and the y-intercept of the $I-V^2$ graph
2. the negative of the ratio of the y-intercept and the slope, and the y-intercept of the $I-V^2$ graph
3. the slope and the y-intercept of the $\sqrt{I}-V$ graph
4. the negative of the ratio of the y-intercept and the slope, and the y-intercept of the $\sqrt{I}-V$ graph

The equation $$I=I_0\left(1-\frac{V}{V_0}\right)^2$$ can be written as $$\sqrt{I}=-\frac{\sqrt{I_0}}{V_0}V+\sqrt{I_0}$$ Hence, $\sqrt{I}-V$ graph is a straight line with slope $-\frac{\sqrt{I_0}}{V_0}$ and y-intercept $\sqrt{I_0}$

Hence, answer is (D)