(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.