17th Feb 2021

# (a)

First compute total energy of system, let $$\gamma$$ be the Lorentz factor of the positron, $E_{\text{total}} = E + m_e c^2 = (\gamma + 1) m_e c^2.$ By conservation of energy, and assuming energy of both photons are the same, $E_{\text{photon}} = \frac{\gamma + 1}{2} m_e c^2.$ Now we can compute the momentum of each photon as $p_{\text{photon}} = \frac{E_{\text{photon}}}{c} = \frac{\gamma + 1}{2} m_e c.$ As the total momentum of the system is $$\gamma m_e v$$ initially, by conservation of momentum (in the horizontal direction), \begin{align*} \gamma m_e v &= 2 \cos\left(\frac{\alpha}{2}\right) p_{\text{photon}} \\ &= 2 \cos\left(\frac{\alpha}{2}\right) \frac{\gamma + 1}{2} m_e c \\ \cos\left(\frac{\alpha}{2}\right) &= \frac{\gamma}{\gamma+1}\frac{v}{c} \\ \alpha &= 2\cos^{-1} \left( \frac{\gamma}{\gamma+1}\frac{v}{c} \right) \\ &= 2\cos^{-1} \left( \frac{v}{c + \sqrt{c^2-v^2}} \right) \end{align*}

# (b)

The minimum energy is given by $\min_{\Delta p, \Delta x} \left[ \frac{(\Delta p)^2}{2m} + \frac12 k(\Delta x)^2 \right]$ where $$\Delta p\Delta x \geq \frac{\hbar}2$$. To achieve minimum, we let equality hold in the uncertainty principle, so we substitute $$\Delta p = \frac{\hbar}{2(\Delta x)}$$ and compute $\min_{\Delta x} \left[ \frac{\hbar^2}{8m(\Delta x)^2} + \frac12 k(\Delta x)^2 \right].$ To minimise, differentiate the expression w.r.t. $$\Delta x$$ $-\frac{\hbar^2}{4m(\Delta x)^3} + k(\Delta x) = 0$ solving, we find that $\Delta x = \sqrt{\frac{\hbar^2}{4mk}}.$ To check that this point is minimum we do second derivative test $\frac{3\hbar^2}{4m(\Delta x)^4} + k > 0$ as the second derivative is always positive we have a minimum, therefore the minimum energy is $\frac{\hbar^2}{8m}\sqrt{\frac{4mk}{\hbar^2}} + \frac12 k\sqrt{\frac{\hbar^2}{4mk}} = \frac{\hbar}{2}\sqrt{\frac{k}{m}}.$

# (c)

Let $$A$$ be surface area of filament, by Stefan-Boltzmann law we have $I = \frac{75}{A} = \sigma T^4$ solving, we have \begin{align*} A &= \frac{75}{5.67\times 10^{-8} \cdot (3000)^4} \\ &= 1.6\times 10^{-5}\ \mathrm{m}^{2}. \end{align*}

# (d)

The maximum kinetic energy $$K_{\max}$$ is given by \begin{align*} K_{\max} &= hf - W \\ &= \frac{hc}{\lambda} - W \\ &= \frac{6.626\times 10^{-34} \cdot 2.998\times 10^{8}}{350\times 10^{-9}}\ \mathrm{J} - 2.24\ \mathrm{eV} \\ &= 5.676\times 10^{-19}\ \mathrm{J} - 2.24\ \mathrm{eV} \\ &= 3.54\ \mathrm{eV} - 2.24\ \mathrm{eV} \\ &= 1.30\ \mathrm{eV} \end{align*}