Problem set 58

1. For an electron moving through a one- dimensional periodic lattice of periodicity a, which of the following corresponds to an energy eigenfunction consistent with Bloch’s theorem?
1. $\psi(x)=A\exp{\left( i\left[\frac{\pi x}{a}+\cos{\left(\frac{\pi x}{2a}\right)}\right]\right)}$
2. ${\scriptstyle\psi(x)=A\exp{\left( i\left[\frac{\pi x}{a}+\cos{\left(\frac{2\pi x}{a}\right)}\right]\right)}}$
3. ${\scriptstyle \psi(x)=A\exp{\left( i\left[\frac{2\pi x}{a}+i\cosh{\left(\frac{2\pi x}{a}\right)}\right]\right)}}$
4. $\psi(x)=A\exp{\left( i\left[\frac{\pi x}{2a}+i\left|\frac{\pi x}{2a}\right|\right]\right)}$

According to Bloch theorem $$\psi(x)=e^{-ikx}u(x)$$ where $\psi(x)$ and $u(x)$ have same periods. Clearly, $\psi(x)=A\exp{\left( i\left[\frac{\pi x}{a}+\cos{\left(\frac{2\pi x}{a}\right)}\right]\right)}$ satisfies above condition.

Hence, answer is (B)

2. A thin metal film of dimension $2\:mm\times2\:mm$ contains $4\times10^{12}$ electrons. The magnitude of the Fermi wavevector of the system, in the free electron approximation, is
1. $2\sqrt{\pi}\times10^{7}\:cm^{-1}$
2. $\sqrt{2\pi}\times10^{7}\:cm^{-1}$
3. $\sqrt{\pi}\times10^{7}\:cm^{-1}$
4. $2\pi\times10^{7}\:cm^{-1}$

In two-dimensional electron gas approximation magnitude of Fermi wave vector is given by $$k^2=2\pi n$$ where, $n$ is electron density. \begin{align*} k&=\sqrt{2\pi n}\\ &=\sqrt{2\pi}\sqrt{\frac{N}{A}}\\ &=\sqrt{2\pi}\sqrt{\frac{4\times10^{12}}{4\times 10^{-2}}}\\ &=\sqrt{2\pi}\times10^{7}\:cm^{-1} \end{align*}

Hence, answer is (B)

3. A positron is suddenly absorbed by the nucleus of a tritium $(^3_1H)$ atom to turn the latter into a $He^+$ ion. If the electron in the tritium atom was initially in the ground state, the probability that the resulting $He^+$ ion will be in its ground state is
1. 1
2. $\frac{8}{9}$
3. $\frac{128}{243}$
4. $\frac{512}{729}$

Both Tritium and $He^+$ are hydrogenic atoms. The ground state wavefunction of hydrogenic atom is given by $$\psi_{0}=R_{10}Y_{00}$$ where $$R_{10}=2\left(\frac{Z}{a_0}\right)^{3/2}e^{-Zr/a_0}$$ $$Y_{00}=\frac{1}{\sqrt{4\pi}}$$ $$a_0=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}$$ For both Tritium and $He^+$ have reduced mass $\mu$ same. Hence, these two differs in $Z$ values. $$\psi_{0_{He^+}}=2\frac{1}{\sqrt{4\pi}}\left(\frac{2}{a_0}\right)^{3/2}e^{-2r/a_0}$$ $$\psi_{0_{Tritium}}=2\frac{1}{\sqrt{4\pi}}\left(\frac{1}{a_0}\right)^{3/2}e^{-r/a_0}$$ According to sudden approximation probability of transition is given by $$P=\left|\left < \psi_{0_{He^+}}| \psi_{0_{Tritium}}\right >\right|^2$$ \begin{align*} &\left < \psi_{0_{He^+}}| \psi_{0_{Tritium}}\right >\\ &={\textstyle 2^2\frac{1}{4\pi}\left(\frac{1}{a_0}\right)^32^{3/2}\int_0^\infty e^{-3r/a_0}4\pi r^2 dr}\\ &={\textstyle \left(\frac{1}{a_0}\right)^32^{7/2}\int_0^\infty r^2 e^{-3r/a_0} dr}\\ &={\scriptstyle \left(\frac{1}{a_0}\right)^32^{7/2}\!\left(\frac{a_0}{3}\right)^3\int_0^\infty \left(\frac{3}{a_0}r\right)^2 e^{-3r/a_0} d\left(\frac{3}{a_0}r\right)}\\ &={\textstyle \left(\frac{1}{a_0}\right)^32^{7/2}\left(\frac{a_0}{3}\right)^32!}\\ &=\frac{2^{9/2}}{27} \end{align*} where we have used formula $\int_0^\infty x^n e^{-x} dx=n!$ $$P=\frac{2^{9}}{27^2}=\frac{512}{729}$$

Hence, answer is (D)

4. The first order diffraction peak of a crystalline solid occurs at a scattering angle of $30^0$ when the diffraction pattern is recorded using an x-ray beam of wavelength 0.15 nm. If the error in measurements of the wavelength and the angle are 0.01 nm and $1^0$ respectively, then the error in calculating the inter- planar spacing will approximately be
1. $1.1\times 10^{-2}\:nm$
2. $1.3\times 10^{-4}\:nm$
3. $2.5\times 10^{-2}\:nm$
4. $2.0\times 10^{-3}\:nm$

$$2d\sin\theta=\lambda$$ $$d=\frac{\lambda}{2\sin\theta}$$ Let $z=\sin\theta$. $$d=\frac{\lambda}{2z}$$ \begin{align*} \Delta z&=\cos\theta\Delta\theta\\ &=\sin{30}\times 1^0\\ &=\frac{\sqrt{3}}{2}\frac{\pi}{180} \end{align*} \begin{align*} \Delta d&=\left|\frac{\lambda}{2z}\right|\sqrt{\left(\frac{\Delta\lambda}{\lambda}\right)^2+\left(\frac{\Delta z}{z}\right)^2}\\ &={\textstyle\left|\frac{0.15}{2\sin{30}}\right|\sqrt{\left(\frac{0.01}{0.15}\right)^2+\left(\frac{\frac{\sqrt{3}}{2}\frac{\pi}{180}}{\sin{30}}\right)^2}}\\ &=\left|0.15\right|\sqrt{\frac{1}{225}+\frac{3\pi^2}{180^2}}\\ &=0.011=1.1\times 10^{-2}\:nm \end{align*}

Hence, answer is (A)

5. The dispersion relation of electrons in a 3-dimensional lattice in the tight binding approximation is given by, $$\epsilon_k=\alpha\cos{k_xa}+\beta\cos{k_ya}+\gamma\cos{k_za}$$ where $a$ is the lattice constant and $\alpha,\beta,\gamma$ are constants with dimension of energy. The effective mass tensor at the corner of the first Brillouin zone $\left(\frac{\pi}{a},\frac{\pi}{a},\frac{\pi}{a}\right)$ is
1. $\frac{\hbar^2}{a^2}\begin{pmatrix}-\frac{1}{\alpha}&0&0\\0&-\frac{1}{\beta}&0\\0&0&\frac{1}{\gamma}\end{pmatrix}$
2. $\frac{\hbar^2}{a^2}\begin{pmatrix}-\frac{1}{\alpha}&0&0\\0&-\frac{1}{\beta}&0\\0&0&-\frac{1}{\gamma}\end{pmatrix}$
3. $\frac{\hbar^2}{a^2}\begin{pmatrix}\frac{1}{\alpha}&0&0\\0&\frac{1}{\beta}&0\\0&0&\frac{1}{\gamma}\end{pmatrix}$
4. $\frac{\hbar^2}{a^2}\begin{pmatrix}\frac{1}{\alpha}&0&0\\0&\frac{1}{\beta}&0\\0&0&-\frac{1}{\gamma}\end{pmatrix}$

Effective mass tensor is given by $$m^*=\begin{pmatrix}m_{xx}&m_{xy}&m_{xz}\\m_{yx}&m_{yy}&m_{yz}\\m_{zx}&m_{zy}&m_{zz}\end{pmatrix}$$ where $$m_{\alpha\beta}=\frac{\hbar^2}{\frac{\partial^2\epsilon_k}{\partial k_\alpha \partial k_\beta}}\quad \alpha ,\beta=x,y,z$$ $$m_{xx}=\frac{\hbar^2}{\frac{\partial^2\epsilon_k}{\partial k_x \partial k_x}}=-\frac{\hbar^2}{a^2}\frac{1}{\alpha}\cos{k_xa}$$ $$\left.m_{xx}\right|_{\frac{\pi}{a}}=\frac{\hbar^2}{a^2}\frac{1}{\alpha}$$ Similarly, find other

Hence, answer is (C)