Problem set 35

1. The energy of formation of a vacancy in copper is $1eV$. The number of vacancies per mole below its melting point $1356^oK$ is:
1. $1.15\times10^{20}$
2. $4\times10^{20}$
3. $2\times10^{20}$
4. $3.30\times10^{20}$
$$\frac{n}{N}=e^{\left(-\frac{E}{k_BT}\right)}$$ $$E=-k_BT\ln{\frac{n}{N}}$$ \begin{align*} E&={\scriptstyle-1.38\times10^{-23}\times1356\times\ln{\frac{1}{6.022140857\times10^{23}}}} &=-6.404} \end{align*}
2. A system is known to be in a state described by the wave function $$\psi(\theta,\phi)=\frac{1}{\sqrt{30}}\left[5Y_4^0+5Y_6^0+25Y_6^3\right]$$ where $Y_l^m$ are spherical harmonics. The probability of finding the system in a state with $m=0$ is :
1. Zero
2. $\frac{6}{\sqrt{30}}$
3. $\frac{6}{30}$
4. $\frac{13}{15}$
\begin{align*} |\psi|^2&=\frac{1}{30}\left[25\left|Y_4^0\right|^2+\left|Y_6^0\right|^2\right]\\ &=\frac{26}{30}=\frac{13}{15} \end{align*} Hence, answer is (D)
3. For attractive one-dimensional delta function potential situated at $x=0$, the wave function of the bound state is given by:
1. $\psi(x)=e^{-\alpha x}$
2. $\psi(x)=e^{-\alpha |x|}$
3. $\psi(x)=e^{-\alpha x^2}$
4. $\psi(x)=\sin{\alpha x}$
The attractive delta function is given by $$V(x)=-a\delta(x)$$ The $\delta$-function is infinite at $x=0$ and zero elsewhere. For bound state total energy $E$ of the particle must be less than the potential energy $V(X)$. Hence, $E$ must be negative where $V(x)=0$. The Schrodinger equation at these points will be $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi\quad\text{ or }\frac{d^2\psi}{dx^2}=\alpha^2\psi$$ where, $\alpha=\sqrt{-\frac{2mE}{\hbar^2}}$. The general solution of above equation can be written as $$\psi(x)=Ae^{-\alpha x}+Be^{\alpha x}$$ For $x<0$, the term, $Ae^{-\alpha x}\rightarrow\infty$ as $x\rightarrow-\infty$, and for $x>0$, the term $Ae^{\alpha x}\rightarrow\infty$ as $x\rightarrow\infty$. Hence, we will have to write separate solutions for these two regions. $$\psi_-(x)=Be^{\alpha x}\quad\text{for} x<0$$ $$\psi_+(x)=Ae^{-\alpha x}\quad\text{for} x>0$$ At $x=0$ these two solutions must be continous. Hence, we get $A=B$. $$\psi_-(x)=Ae^{\alpha x}\quad\text{for} x<0$$ $$\psi_+(x)=Ae^{-\alpha x}\quad\text{for} x>0$$ Thus, $$\psi(x)=Ae^{-\alpha |x|}\quad\text{for all values of } x$$ Hence, answer is (B)
4. A one-dimensional simple harmonic oscillator with generalized coordinate $q$ is subjected to an extra additional potential energy of the form $$V(t)=q^2t+q\dot qt^2$$ The Lagrange's equation of the oscillator due to the extra potential will contain:
1. an extra term proportional to $t$
2. an extra term proportional to $t^2$
3. an extra term proportional to $(t+t^2)$
4. no extra term
$$L=\frac{1}{2}m\dot q^2-\frac{1}{2}kq^2-q^2t-q\dot qt^2$$ $$\frac{\partial L}{\partial \dot q}=m\dot q-qt^2$$ $$\frac{\partial L}{\partial q}=-kq-2qt-\dot qt^2$$ \begin{align*} &{\scriptstyle\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)-\frac{\partial L}{\partial q}=\frac{d}{dt}\left(m\dot q-qt^2\right)-\left(-kq-2qt-\dot qt^2\right)}\\ &=m\ddot q-\dot qt^2-2qt+kq+2qt+\dot qt^2\\ &=m\ddot q+kq=0 \end{align*} Hence, there is no change in Lagrange's equation of motion. Hence, answer is (D)
5. The variational method in perturbation theory, when applied to obtain the value of the ground state energy:
1. Always gives exact ground state energy
2. gives energy value lower than the exact ground state energy
3. gives energy value which is sometimes higher than or sometimes lower than the exact ground state energy
4. gives energy value higher than or equal to the exact ground state energy
Variational method gives energy value higher than or equal to the exact ground state energy. Hence, answer is (D).