24th March 2021

# (a)

As initial radius was $$a_0 = 0.529\times 10^{-10}\ \mathrm{m}$$, the total energy of the atom is given by $E = -\frac{e^2}{8\pi\epsilon_0 r(t)}$ by conservation of energy, $$\frac{dE}{dt} = -P$$, we can compute $\frac{dE}{dt} = \frac{e^2}{8\pi\epsilon_0 r(t)^2} r'(t)$ note that in the expression for radiated power, $\left\lvert \vec{w} \right\rvert = r''(t)$ so we can combine our equations \begin{align*} \frac{e^2}{8\pi\epsilon_0 r(t)^2} r'(t) &= -\frac{e^2}{6\pi\epsilon_0 c^3} r''(t)^2 \\ \frac{1}{4 r(t)^2} r'(t) &= -\frac{1}{3 c^3} r''(t)^2 \end{align*} rearranging, $r'(t) = -\frac43 \frac{r(t)^2 r''(t)^2}{c^3}.$ Classically we can express $$r''(t)$$ in terms of $$r$$ by Coulomb’s law as $r''(t) = \frac{F}{m} = \frac{e^2}{4\pi\epsilon_0 m r^2}$ and rewrite our equation as $r'(t) = -\frac{e^4}{12 \pi^2 \epsilon_0^2 m^2 c^3} r(t)^{-2}.$ Now the time is given by \begin{align*} t &= \int_{a_0}^0 \frac{dt}{dr}\ dr \\ &= -\frac{12 \pi^2 \epsilon_0^2 m^2 c^3}{e^4} \int_{a_0}^0 r^2\ dr \\ &= -\frac{4 \pi^2 \epsilon_0^2 m^2 c^3}{e^4} \left[ r^3 \right]^0_{a_0} \\ &= \frac{4 \pi^2 \epsilon_0^2 m^2 c^3}{e^4} a_0^3 \\ &= 1.55\times 10^{-11}\ \mathrm{s}. \end{align*}

# (b)

Assume that the mean value of kinetic energy is quantised as $K = \frac{nh f_{\text{orb}}}2.$ $$K$$ is related to $$L_z$$ by the relation $L_z = rmv; K = \frac12 mv^2 \implies L_z = \frac{2Kr}{v}.$ Combining expressions we obtain $L_z = \frac{2r}{v} \frac{nh f_{\text{orb}}}2 = \frac{rnh}{v} f_{\text{orb}}$ but as $$f_{\text{orb}} = \frac1T$$ and the period $$T = \frac{2\pi r}{v}$$, we have $f_{\text{orb}} = \frac{v}{2\pi r}$ and substituting into expression for $$L_z$$ gives us $L_z = \frac{rnh}{v} \frac{v}{2\pi r} = \frac{nh}{2\pi} = n\hbar.$

# (c)

The energy difference is given by $\Delta E = 2 \mu_B B = 1.41\times 10^{-23}\ \mathrm{J}.$

# (d)

$f = \frac{c}{\lambda} = \frac{3Rc}{4} \left( Z - 1 \right)^2$ computing, we have $$Z - 1 = 25$$, so $$Z = 26$$ and the element is iron.