Problem Set 37

1. The LS configurations of the ground state of $^{12}Mg$, $^{13}Al$, $^{17}Cl$ and $^{18}Ar$ are, respectively,
1. $^3S_1$, $^2P_{1/2}$, $^2P_{1/2}$ and $^1S_{0}$
2. $^3S_1$, $^2P_{3/2}$, $^2P_{3/2}$ and $^3S_{1}$
3. $^1S_0$, $^2P_{1/2}$, $^2P_{3/2}$ and $^1S_{0}$
4. $^1S_0$, $^2P_{3/2}$, $^2P_{1/2}$ and $^3S_{1}$

Electronic configuration of $^{12}Mg$ is $1s^22s^22p^63s^2$

There are two 3s electrons with $s_1=\frac{1}{2}$ and $s_2=\frac{1}{2}$. Hence, $S=1$ and $2S+1=3$

For two 3s electrons each with $l=0$, $m_l=0$. Hence, $L=0$

$J=L+S=0$. Hence, term symbol for $^{12}Mg$ is $^1S_0$

For, $^{13}Al$, electronic configuration is $1s^22s^22p^63s^23p^1$

$S=\frac{1}{2}$, $2S+1=2$, $L=1$, $J=L-S=1-\frac{1}{2}=\frac{1}{2}$ (less than half-filled)

Hence, term symbol for $^{13}Al$ is $^2P_{1/2}$

Hence, answer is (C).

2. The frequency dependent dielectric constant of a material is given by $$\epsilon(\omega)=1+\frac{A}{\omega_0^2-\omega^2-i\omega\gamma}$$ where $A$ is a positive constant, $\omega_0$ the resonant frequency and $\gamma$ the damping coefficient. For an electromagnetic wave of angular frequency $\omega < < \omega_0$, which of the following is true? (Assume that $\frac{\gamma}{\omega_0} < < 1$.)
1. There is negligible absorption of the wave
2. The wave propagation is highly dispersive
3. There is strong absorption of the electromagnetic wave
4. The group velocity and the phase velocity will have opposite sign
\begin{align*} &\epsilon(\omega)=1+\frac{A}{\omega_0^2-\omega^2-i\omega\gamma}\\ &={\scriptstyle1+\frac{A\left(\omega_0^2-\omega^2+i\omega\gamma\right)}{\left(\omega_0^2-\omega^2+i\omega\gamma\right)\left(\omega_0^2-\omega^2-i\omega\gamma\right)}}\\ &={\scriptstyle1+\frac{A\left(\omega_0^2-\omega^2+i\omega\gamma\right)}{\left(\omega_0^2-\omega^2\right)^2+\omega^2\gamma^2}}\\ &={\scriptstyle1+\frac{A\left(\omega_0^2-\omega^2\right)}{\left(\omega_0^2-\omega^2\right)^2+\omega^2\gamma^2}+i\frac{A\omega\gamma}{\left(\omega_0^2-\omega^2\right)^2+\omega^2\gamma^2}}\\ \end{align*} Fro $\omega < < \omega_0$, imaginary part of $\epsilon(\omega)$ is negligibly small. Hence, there is negligible absorption of the wave.

Hence, answer is (A)

3. The state diagram corresponding to the following circuit is
   

In options, ⓪ and ① represent states of Flip-Flop i.e. output $A$. Suppose flip-flop is in state ⓪ i.e. $A=0$. If the $x, y$ inputs are $0,0$ then output $A$ will become 1 i.e. state will change from ⓪ to ①

Now, suppose the flip-flop is in state ⓪ i.e. $A=0$. If the $x, y$ inputs are $1,1$ then output $A$ will remain 0 i.e. the state remains same ⓪.

Hence, option (D) is correct

4. Consider a random walker on a square lattice. At each step the walker moves to a nearest neighbour site with equal probability for each of the four sites. The walker starts at the origin and takes 3 steps. The probability that during this walk no site is visited more than once is
1. 12/27
2. 27/64
3. 3/8
4. 9/16

During first step of random walk the probability that during this walk no site is visited more than once is 1

During second step of random walk the probability that during this walk no site is visited more than once is 3/4, because there are 3 sites which are unvisited and each site has probability 1/4

Similarly, during third step of random walk the probability is 3/4

Hence, the probability that during first, second and third step, no site is visited more than once is $$p=1\times\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}$$

Hence, answer is (D)

5. A canonical transformation $(p,q)\rightarrow(P,Q)$ is performed on the Hamiltonian $H=\frac{1}{2m}p^2+\frac{1}{2}m\omega^2q^2$ via the generating function $F=\frac{1}{2}m\omega q^2\cot{Q}$. If $Q(0)=0$, which of the following graphs shows schematically the dependence of $Q(t)$ on $t$?
1. 
2. 

4. 
$H=\frac{1}{2m}p^2+\frac{1}{2}m\omega^2q^2$, $F=\frac{1}{2}m\omega q^2\cot{Q}$ $$H'=H-\frac{\partial F}{\partial t}=H$$ For the generating function $F$ we have $\frac{\partial F}{\partial q}=p$ and $\frac{\partial F}{\partial Q}=-P$

$\frac{\partial F}{\partial q}=p$ gives $$m\omega q\cot{Q}=p$$

$\frac{\partial F}{\partial Q}=-P$ gives $$-\frac{1}{2}m\omega q^2 \text{ cosec}^2Q=-P$$ $$q^2=\frac{2P}{m\omega \text{ cosec}^2Q}$$ \begin{align*} H'&={\scriptstyle\frac{1}{2m}m^2\omega^2q^2\cot^2Q+\frac{1}{2}m\omega^2\frac{2P}{m\omega \text{ cosec}^2Q}}\\ &={\scriptstyle\frac{1}{2m}m^2\omega^2\frac{2P}{m\omega \text{ cosec}^2Q}\cot^2Q+\frac{1}{2}m\omega^2\frac{2P}{m\omega \text{ cosec}^2Q}}\\ &=P\omega\cos^2Q+P\omega\sin^2Q\\ &=P\omega \end{align*} Using canonical equation $\dot Q=\frac{\partial H'}{\partial P}$ we have $$\frac{dQ}{dt}=\omega$$ $$Q=\omega t+c$$ where $c$ in integrating constant. But at $t=0$, $Q(t)=0$, hence, $c=0$ $$Q=\omega t$$ Which is equation of straight line.

Hence, answer is (D)