Problem set 56

1. A system of $N$ distinguishable particles, each of which can be in one of the two energy levels $0$ and $\epsilon$, has a total energy $n\epsilon$, where $n$ is an integer. The entropy of the system is proportional to
1. $N \ln{n}$
2. $n \ln{N}$
3. $\ln{\frac{N!}{n!}}$
4. $\ln{\left(\frac{N!}{n!(N-n)!}\right)}$

Let us consider three distinguishable particles A, B, C to be arranged in two energy levels $0$ and $\epsilon$, such that total energy is $n\epsilon$ ($n=0,1,3$). The total number of microstates are given as follow:

Total energy $n\epsilon$	0	$\epsilon$	Number of microstates
$0\epsilon$	ABC	-	1
$1\epsilon$	AB	C	3
	AC	B
	BC	A
$2\epsilon$	A	BC	3
	B	AC
	C	AB
$3\epsilon$	-	ABC	1

In general, the number microstates for $N$ particles for total energy $n\epsilon$ is given by $$\Omega=\frac{N!}{n!(N-n)!}$$ Hence, entropy of system is given by $$S=k_B\ln\Omega=k_B\ln{\left(\frac{N!}{n!(N-n)!}\right)}$$

Hence, answer is (D)

2. The wavefunction of a particle in one-dimension is denoted by $\psi(x)$ in the coordinate representation and $\phi(p)=\int \psi(x) e^{-ipx/\hbar}\:dx$ in the momentum representation. If the action of an operator $\hat T$ on $\psi(x)$ is given by $\hat T\psi(x)=\psi(x+a)$, where $a$ is constant, then $\hat T\phi(p)$ is given by
1. $-\frac{i}{\hbar}ap\phi(p)$
2. $e^{-iap/\hbar}\phi(p)$
3. $e^{+iap/\hbar}\phi(p)$
4. $\left(1+\frac{i}{\hbar}ap\right)\phi(p)$

$$\phi(p)=\int \psi(x) e^{-ipx/\hbar}\:dx$$ \begin{align*} \hat T\phi(p)&=\int \hat T\psi(x) e^{-ipx/\hbar}\:dx\\ &=\int \psi(x+a) e^{-ipx/\hbar}\:dx\\ &=\!\int\!\psi(x+a) e^{\frac{-ip(x+a)}{\hbar}}e^{\frac{+ipa}{\hbar}}dx\\ &=e^{\frac{+ipa}{\hbar}}\int\! \psi(x+a) e^{\frac{-ip(x+a)}{\hbar}}dx\\ &=e^{+ipa/\hbar}\phi(p) \end{align*}

Hence, answer is (C)

3. A proton moves with a speed of 300m/s in a circular orbit in the xy-plane in a magnetic 1 tesla along the positive z-direction. When an electric field of 1 V/m is applied along the positive y-direction, the center of the circular orbit
1. remains stationary
2. moves at 1 m/s along the negative x-direction
3. moves at 1 m/s along the positive z-direction
4. moves at 1 m/s along the positive x-direction

The addition of an electric field perpendicular to a given magnetic field simply causes the particle to drift perpendicular to both the electric and magnetic field with the fixed velocity $$\vec v_{EB}=\frac{\vec E\times\vec B}{B^2}$$ In the present case, $\vec E=1\hat j$ and $\vec B=1\hat k$ $$\vec v_{EB}=\frac{\hat j\times\hat k}{1}=\hat i$$ Hence, center of circular moves at 1 m/s along the positive x-direction

Hence, answer is (D)

4. Suppose the yz-plane forms a chargeless boundary between two media of permittivities $\epsilon_{left}$ and $\epsilon_{right}$ where $\epsilon_{left}:\epsilon_{right}=1:2$. If the uniform electric field on the left is $\vec E_{left}=c\left(\hat i+\hat j+\hat k\right)$ (where $c$ is constant), then the electric field on the right $\vec E_{right}$ is
1. $c\left(2\hat i+\hat j+\hat k\right)$
2. $c\left(\hat i+2\hat j+2\hat k\right)$
3. $c\left(\frac{1}{2}\hat i+\hat j+\hat k\right)$
4. $c\left(\hat i+\frac{1}{2}\hat j+\frac{1}{2}\hat k\right)$

Here the electric fields in two media should satisfy the condition $$ E_{left}\epsilon_{left}\cos\theta_1= E_{right}\epsilon_{right}\cos\theta_2$$ where, $\theta_1$ is the angle between $\vec E_{left}$ and unit normal to yz-plane in left medium and $\theta_2$ is the angle between $\vec E_{right}$ and unit normal to yz-plane in right medium. $$\vec E_{left}=c\left(\hat i+\hat j+\hat k\right)$$ The unit vector normal to yz-plane in left medium is $\hat n=-\hat i$. Hence, angle between $\vec E_{left}$ and $\hat n$ is given by \begin{align*} \cos\theta_1&=\frac{\left|\vec E_{left}\cdot\hat n\right|}{\left|\vec E_{left}\right|\left|\hat n\right|}\\ &=\frac{c}{c\sqrt{3}\times 1}\\ \cos\theta_1&=\frac{1}{\sqrt{3}} \end{align*} \begin{align*} E_{right}\cos\theta_2&= E_{left}\frac{\epsilon_{left}}{\epsilon_{right}}\cos\theta_1\\ &=c\sqrt{3}\times\frac{1}{2}\times\frac{1}{\sqrt{3}}\\ E_{right}\cos\theta_2&=\frac{c}{2} \end{align*} The vector which satisfies above condition is $\vec E_{right}=c\left(\frac{1}{2}\hat i+\hat j+\hat k\right)$

Hence, answer is (C)

5. A particle of mass $2\: kg$ is moving such that at time $t$, its position, in metre, is given by $\vec r(t) = 5\hat i- 2t^2\hat j$. The angular momentum of the particle at $t = 2\: s$ about the origin, in $kg\: m^2\: s^{-1}$, is
1. $-40\hat k$
2. $-80\hat k$
3. $80\hat k$
4. $40\hat k$

$$\vec r(t) = 5\hat i- 2t^2\hat j$$ $$\vec v(t) = - 4t\hat j$$ \begin{align*} \vec L&=m\left(\vec r\times\vec v\right)\\ &=m\begin{vmatrix}\hat i&\hat j&\hat k\\5&- 2t^2&0\\0&-4t&0\end{vmatrix}\\ &=-20mt\hat k=-80\hat k \end{align*}

Hence, answer is (B)