Problem set 36

1. For a finite square well potential in one dimension:
1. It is possible that no bound state exits
2. There is always at least one bound state
3. Bound states have degeneracy = 2
4. Energy levels of bound states are equally spaced
(B) There is always at least one bound state
2. A particle with spin $\frac{1}{2}$ is in state with eigenstate of $S_z$. Then the expectation values of $S_x$, $S_x^2$ in this state are given by:
1. $-\frac{\hbar}{2}$, $\frac{1}{4}\hbar$
2. $0$, $\frac{3}{4}\hbar^2$
3. $\frac{\hbar}{2}$, $\frac{3}{4}\hbar^2$
4. $0$, $\frac{1}{4}\hbar^2$
For spin $\frac{1}{2}$ particle the eigenstates are given by $\chi_1=\begin{bmatrix}1\\0\end{bmatrix}$ and $\chi_2=\begin{bmatrix}0\\1\end{bmatrix}$. The spin components $\hat S_x$, $\hat S_y$ and $\hat S_z$ are given by $$\hat S_x=\frac{\hbar}{2}\sigma_x=\frac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}$$ $$\hat S_y=\frac{\hbar}{2}\sigma_y=\frac{\hbar}{2}\begin{bmatrix}0&-i\\i&0\end{bmatrix}$$ $$\hat S_z=\frac{\hbar}{2}\sigma_z=\frac{\hbar}{2}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$ \begin{align*} \left < \hat S_x\right > &=\left < \chi_1\left|\hat S_x\right|\chi_1\right > \\ &=\begin{bmatrix}1&0\end{bmatrix}\frac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\\ &=0 \end{align*} \begin{align*} \left < \hat S_x^2\right > &=\left < \chi_1\left|\hat S_x\hat S_x\right|\chi_1\right > \\ &{\scriptstyle=\begin{bmatrix}1&0\end{bmatrix}\frac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\frac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}}\\ &=\frac{1}{4}\hbar^2 \end{align*} Hence, answer is (D)
3. The differential cross-section for a central potential is equal to
1. $f(\theta,\phi)$
2. $f^*(\theta,\phi)$
3. $f^*(\theta,\phi)f(\theta,\phi)$
4. $|f(\theta,\phi)|$
where asymptotic form of the wave function of the relative motion is given by: $$A\left[e^{ikz}+\frac{f(\theta,\phi)}{r}e^{ikr}\right]$$
Differential cross-section is given by $\frac{d\sigma}{d\Omega}=|f(\theta,\phi)|^2$. Hence, answer is (C).
4. The de Broglie wavelength of a helium atom at 300 K is 0.06 $A^o$. The de Broglie wavelength of neon atom (5 times heavier than helium) at 600 K will be:
1. 6 Å
2. 0.06 Å
3. $0.06\times\sqrt{10}$Å
4. $\frac{0.06}{\sqrt{10}}$Å
Thermal de Broglie wavelength is given by $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$. In the nonrelativistic case the effective kinetic energy of free particles is $E = \pi k_B T $. $$\lambda=\frac{h}{\sqrt{2\pi mk_BT}}$$ $$\lambda_{He}\propto\frac{1}{\sqrt{m_{He}T}}$$ and$$\lambda_{Ne}\propto\frac{1}{\sqrt{m_{Ne}T}}$$ $$\frac{\lambda_{Ne}}{\lambda_{He}}=\frac{\sqrt{m_{He}T_{He}}}{\sqrt{m_{Ne}T_{Ne}}}$$ $$\lambda_{Ne}=\sqrt{\frac{m_{He}T_{He}}{m_{Ne}T_{Ne}}}\lambda_{He}$$ $\lambda_{Ne}=\sqrt{\frac{1\times 300}{5\times600}}0.6=\frac{0.6}{\sqrt{10}}$Å
Hence, answer is (D)
5. If a charged particle $q$ moves along a circle of radius $r=100mm$ in a uniform magnetic field $B=10mT$, then the period of revolution of the particle $(m_p=1.67\times10^{-27}kg, q=1.6\times10^{-19}C)$
1. 6.55 ms
2. 6.55 $\mu$s
3. 6.55 ns
4. 3$\mu$s
For circular motion radial force is equal to force due to magnetic field. $$\frac{mv^2}{r}=qvB$$ $$T=\frac{2\pi}{\omega}=\frac{2\pi r}{v}=\frac{2\pi m}{qB}$$ \begin{align*} T&=\frac{2\times3.14\times1.67\times10^{-27}}{1.6\times10^{-19}\times10\times10^{-3}}\\ &=6.55\mu s \end{align*}