Problem set 85

1. In a normal Zeeman effect experiment using a magnetic field of strength 0.3 T, the splitting between the components of a 660 nm spectral line is
1. 12 pm
2. 10 pm
3. 8 pm
4. 6 pm

$$\Delta\lambda=\frac{\mu_BB\lambda^2}{hc}$$ $$\Delta\lambda={\scriptstyle \frac{9.274\times10^{-24}\times0.3\times(660\times10^{-9})^2}{6.626\times10^{-34}\times3\times10^8}}$$ $$\Delta\lambda=6 pm$$

Hence, answer is (D)

2. What is the Fourier transform $\int dxe^{ikx}f(x)$ of $f(x)=\delta(x)+\sum\limits_{n=1}^\infty\frac{d^n}{dx^n}\delta(x)$, where $\delta(x)$ is the Dirac delta-function?
1. $\frac{1}{1-ik}$
2. $\frac{1}{1+ik}$
3. $\frac{1}{k+i}$
4. $\frac{1}{k-i}$

Using identity $\int dx f(x)\delta^{(n)}(x-y)=(-1)^nf^{(n)}(y)$, where $f^{(n)}(y)$ is $n^{th}$ derivative of $f$. $$\begin{align*} &\int\!\! dxe^{ikx}\!f(x)\!=\!{\scriptstyle \int dxe^{ikx}\left(\delta(x)+\sum\limits_{n=1}^\infty\frac{d^n}{dx^n}\delta(x)\right)}\\ &={\scriptstyle \int dxe^{ikx}\delta(x)+\int dxe^{ikx}\sum\limits_{n=1}^\infty\frac{d^n}{dx^n}\delta(x)}\\ &={\scriptstyle 1+\int dxe^{ikx}\left(\delta^{(1)}(x)+\delta^{(2)}(x)+\delta^{(3)}(x)+\dots\right)}\\ &={\scriptstyle 1+(-1)^1ik+(-1)^2(ik)^2+(-1)^3(ik)^3+\cdots}\\ &={\scriptstyle 1-ik-k^2+ik^3+k^4-ik^5\cdots}\\ &={\scriptstyle \left(1-k^2+k^4-k^6+\cdots\right)-ik+ik^3-ik^5\cdots}\\ &={\scriptstyle \left(1-k^2+k^4-k^6+\cdots\right)-ik\left(1-k^2+k^4-k^6+\cdots\right)}\\ &={\scriptstyle \left(1-k^2+k^4-k^6+\cdots\right)(1-ik)}\\ &={\scriptstyle \left(1+(ik)^2+(ik)^4+(ik)^6+\cdots\right)(1-ik)}\\ &=\frac{1}{1-(ik)^2}(1-ik)\\ &=\frac{1}{1+ik} \end{align*}$$

Hence, answer is (B)

3. A canonical transformation $(q,p)\rightarrow (Q,P)$ is made through the generating function $F(q,P)=q^2P$ on the Hamiltonian $H(q,p)=\frac{p^2}{2\alpha q^2}+\frac{\beta}{4}q^4$ where $\alpha$ and $\beta$ are constants. The equations of motion for $(Q,P)$ are
1. $\dot Q=P/\alpha$ and $\dot P=-\beta Q$
2. $\dot Q=4P/\alpha$ and $\dot P=-\beta Q/2$
3. $\dot Q=P/\alpha$ and $\dot P=-\frac{2P^2}{Q}-\beta Q$
4. $\dot Q=2P/\alpha$ and $\dot P=-\beta Q$

The generating function may be any of the following types: $F_1(q,Q)$, $F_2(q,P)$, $F_3(p,Q)$, $F_4(p,P)$. For these generating functions we have following relations between generating function and canonical variables

Generating function	Relations
$F_1(q,Q)$	$p_i=\frac{\partial F_1}{\partial q_i}$, $P_i=-\frac{\partial F_1}{\partial Q_i}$, $K=H+\frac{\partial F_1}{\partial t}$
$F_2(q,P)$	$p_i=\frac{\partial F_2}{\partial q_i}$, $Q_i=\frac{\partial F_2}{\partial P_i}$, $K=H+\frac{\partial F_2}{\partial t}$
$F_3(p,Q)$	$q_i=-\frac{\partial F_3}{\partial p_i}$, $P_i=-\frac{\partial F_3}{\partial Q_i}$, $K=H+\frac{\partial F_3}{\partial t}$
$F_4(p,P)$	$q_i=-\frac{\partial F_4}{\partial p_i}$, $Q_i=\frac{\partial F_4}{\partial P_i}$, $K=H+\frac{\partial F_4}{\partial t}$

In the present problem generating function $F(q,P)=q^2P$ is of second kind. Hence, $$p=\frac{\partial F}{\partial q}=2qP$$ $$Q=\frac{\partial F}{\partial P}=q^2$$ $$\begin{align*} K&=H+\frac{\partial F_2}{\partial t}=H\\ &=\frac{p^2}{2\alpha q^2}+\frac{\beta}{4}q^4\\ K&=\frac{2P^2}{\alpha}+\frac{\beta}{4}Q^2\\ \end{align*}$$ Hence, equations of motions are $$\dot P=-\frac{\partial K}{\partial Q}=-\frac{\beta}{2}Q$$ $$\dot Q=\frac{\partial K}{\partial P}=\frac{4P}{\alpha}$$

Hence, answer is (B)

4. The internal energy $E(T)$ of a system at a fixed volume is found to depend on the temperature $T$ as $E(T)=aT^2+bT^4$. Then the entropy $S(T)$, as a function of temperature, is
1. $\frac{1}{2}aT^2+\frac{1}{4}bT^4$
2. $2aT^2+4bT^4$
3. $2aT+\frac{4}{3}bT^3$
4. $2aT+2bT^3$

$$dU=TdS-PdV$$ At constant volume $PdV=0$ $$dU=TdS$$ $$\begin{align*} S&=\int\frac{1}{T}dU\\ &=\int\frac{1}{T}dE\\ &=\int\frac{1}{T}\left(2aT\:dT+4bT^3dT\right)\\ &=\int\left(2a\:dT+4bT^2dT\right)\\ &=2aT+\frac{4}{3}bT^3 \end{align*}$$

Hence, answer is (C)

5. Consider a gas of Cs atoms at a number density of $10^{12}$ atoms/cc. When the typical inter-particle distance is equal to the thermal de Broglie wavelength of the particles, the temperature of the gas is nearest to (Take the mass of a Cs atom to be $22.7\times10^{-26}$ kg.)
1. $1\times10^{-9}$ K
2. $7\times10^{-5}$ K
3. $1\times10^{-3}$ K
4. $2\times10^{-8}$ K

Density of particles $n=\frac{N}{V}$. Hence, inter-particle separation is given by $$d=\left(\frac{V}{N}\right)^{\frac{1}{3}}=\left(\frac{1}{n}\right)^{\frac{1}{3}}$$ $$d=\left(\frac{1}{10^{12}}\right)^{\frac{1}{3}}=10^{-4}cm=10^{-6}m$$ $$\lambda_{th}=10^{-6}m$$ $$\lambda_{th}=\frac{h}{\sqrt{2\pi mk_BT}}$$ $$\begin{align*} T&=\frac{h^2}{2\pi mk_B\lambda_{th}^2}\\ &={\scriptstyle \frac{(6.626\times10^{-34})^{2}}{2\times3.14\times22.7\times10^{-26}\times 1.381 \times 10^{-23}\times10^{-12}}}\\ &=2.23009664\times10^{-8} K \end{align*}$$

Hence, answer is (D)