Problem set 30

1. A particle is confined to the region $x \ge 0$ by a potential which increases linearly as $u(x) = u_0x$. The mean position of the particle at temperature $T$ is
1. $\frac{k_BT}{u_0}$
2. $\frac{(k_BT)^2}{u_0}$
3. $\sqrt{\frac{k_BT}{u_0}}$
4. $u_0k_BT$
$$ < x > = \frac{\int xe^{-\beta H}d\tau}{\int e^{-\beta H}d\tau}$$ where $\beta=\frac{1}{k_BT}$, and $H=\frac{p^2}{2m}+u(x)=\frac{p_x^2}{2m}+u_0x$ \begin{align*} &< x > =\frac{\int xe^{-\beta \left(\frac{p_x^2}{2m}+u_0x\right)}dxdp_x}{\int e^{-\beta \left(\frac{p_x^2}{2m}+u_0x\right)}dxdp_x}\\ &=\frac{\int_0^\infty xe^{-\beta u_0x}dx}{\int_0^\infty e^{-\beta u_0x}dx}\quad{\scriptstyle \int\limits_{0}^\infty x^{n}e^{-\alpha x}dx=\frac{n!}{\left(\alpha\right)^{n+1}}}\\ &=\frac{\frac{1}{(\beta u_0)^2}}{\frac{1}{\beta u_0}}={\frac{1}{\beta u_0}}={\frac{k_BT}{ u_0}} \end{align*} Hence, answer is (A).
2. A plane electromagnetic wave is propagating in a loss-less dielectric. The electric field is given by $${\scriptstyle\vec E(x,y,z,t)=E_0(\hat x+A\hat z)\exp{\left[ik_0\left\{-ct+\left(x+\sqrt{3}z\right)\right\}\right]}}$$ where $c$ is the speed of light in vacuum, $E_0$ , $A$ and $k_0$ are constants and $\hat x$ and $\hat z$ are unit vectors along the x- and z-, axes. The relative dielectric constant of the medium, $\epsilon_r$ and the constant $A$ are
1. $\epsilon_r=4$ and $A=-\frac{1}{\sqrt{3}}$
2. $\epsilon_r=4$ and $A=+\frac{1}{\sqrt{3}}$
3. $\epsilon_r=4$ and $A=\sqrt{3}$
4. $\epsilon_r=4$ and $A=-\sqrt{3}$
General form of wave equation is $\vec E(x,y,z,t)=E_0\hat ne^{i\left(\vec k\cdot\vec r-\omega t\right)}$. Comparing with this equation, we get $$\hat n=(\hat x+A\hat z)$$ \begin{align*} &\vec k\cdot\vec r=k_0\left(x+\sqrt{3}z\right)\\ &\Rightarrow\vec k=k_0\left(\hat x+\sqrt{3}\hat z\right) \end{align*} $$\omega=k_0c$$ $$v=\frac{\omega}{k}=\frac{k_0c}{k_0^2+3k_0^2}=\frac{c}{2}$$ $$\text{Refractive index } n=\frac{c}{v}=2$$ $$\text{Refractive index } n=\sqrt{\epsilon_2}\Rightarrow\epsilon_r=4$$ \begin{align*} &\text{Since }\vec k\cdot\hat n=0\\ &\Rightarrow k_0\left(\hat x+\sqrt{3}\hat z\right)\cdot(\hat x+A\hat z)\\ &=0 \end{align*} $$k_0\left(1+\sqrt{3}A\right)=0\Rightarrow A=-\frac{1}{\sqrt{3}}$$ Hence, answer is (A)
3. In a system consisting of two spin-$\frac{1}{2}$ particles labeled 1 and 2, let $\vec S^{(1)} = \frac{\hbar}{2}\vec\sigma^{(1)}$ and $\vec S^{(2)} = \frac{\hbar}{2}\vec\sigma^{(2)}$ denote the corresponding spin operators. Here $\vec\sigma\equiv (\sigma_x ,\sigma_y ,\sigma_z)$ and $\sigma_x ,\sigma_y ,\sigma_z$ are the three Pauli matrices.
1. In the standard basis the matrices for the operators $ S^{(1)}_x S^{(2)}_y$ and $S^{(1)}_y S^{(2)}_x$ are, respectively,
1. ${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad\frac{\hbar^2}{4}\begin{pmatrix}-1&0\\0&1\end{pmatrix}}$
2. ${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}i&0\\0&-i\end{pmatrix},\quad\frac{\hbar^2}{4}\begin{pmatrix}-i&0\\0&i\end{pmatrix}}$
3. $${\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}},$$$${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}}$$
4. $${\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&-i\\0&0&i&0\end{pmatrix}},$$$${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&-i&0&0\\i&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}}$$
2. These two operators satisfy the relation
1. ${\scriptstyle\left\{S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right\}=S^{(1)}_z S^{(2)}_z}$
2. ${\scriptstyle\left\{S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right\}=0}$
3. ${\scriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=iS^{(1)}_z S^{(2)}_z}$
4. ${\scriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=0}$
This is a double spin system, represented by $4\times4$ matrix representation. \begin{align*} S^{(1)}_x S^{(2)}_y=\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\otimes\frac{\hbar}{2}\begin{pmatrix}0&-i\\i&0\end{pmatrix} \end{align*} where, $\otimes$ represents tensor product. \begin{align*} &S^{(1)}_x S^{(2)}_y={\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0\begin{pmatrix}0&-i\\i&0\end{pmatrix}&1\begin{pmatrix}0&-i\\i&0\end{pmatrix}\\1\begin{pmatrix}0&-i\\i&0\end{pmatrix}&0\begin{pmatrix}0&-i\\i&0\end{pmatrix}\end{pmatrix}}\\ &= \frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix} \end{align*} \begin{align*} S^{(1)}_y S^{(2)}_x&=\frac{\hbar}{2}\begin{pmatrix}0&-i\\i&0\end{pmatrix} \otimes\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\\ &=\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix} \end{align*} Hence, answer is (C) \begin{align*} &{\scriptscriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=S^{(1)}_x S^{(2)}_y\:S^{(1)}_y S^{(2)}_x-S^{(1)}_y S^{(2)}_x\:S^{(1)}_x S^{(2)}_y}\\ &{\scriptscriptstyle=\frac{\hbar^4}{16}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}}\\ &{\scriptscriptstyle-\frac{\hbar^4}{16}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}}\\ &={\scriptscriptstyle\frac{\hbar^4}{16}\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&1\end{pmatrix}-\frac{\hbar^4}{16}\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&1\end{pmatrix}}\\ &=0 \end{align*} Hence, answer is (D)
4. The radius of $^{64}_{29}Cu$ nucleus is measured to be $4.8\times10^{-13} cm$.
1. The radius of $^{27}_{12}Mg$ nucleus can be estimated to be
1. $2.86\times10^{-13} cm$
2. $5.2\times10^{-13} cm$
3. $3.6\times10^{-13} cm$
4. $8.6\times10^{-13} cm$
2. The root-mean square (rms)- energy of a nucleon in a nucleus of atomic number $A$ in its ground state varies as:
1. $A^{4/3}$
2. $A^{1/3}$
3. $A^{-1/3}$
4. $A^{-2/3}$
Nuclear radius is given by $R=R_0A^{1/3}$ $$R_{Cu}=R_0A_{Cu}^{1/3}\quad \text{and}\quad R_{Mg}=R_0A_{Mg}^{1/3}$$ $$\frac{R_{Mg}}{R_{Cu}}=\left(\frac{A_{Mg}}{A_{Cu}}\right)^{1/3}=\left(\frac{27}{64}\right)^{1/3}=\frac{3}{4}$$ \begin{align*} R_{Mg}&=\frac{3}{4}R_{Cu}\\ &=\frac{3}{4}\times4.8\times10^{-13}\\ &=3.6\times10^{-13} cm \end{align*} Hence, answer is (C)
The root-mean square (rms)- energy of a nucleon in a nucleus of atomic number $A$ in its ground state varies as $A^{-1/3}$
Hence, answer is (C)