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(A) The energy of a state is given by $E_n = -\frac{13.6 Z^2}{n^2}$. Let $E_1 = -13.6 Z^2$.The most energetic photon corresponds to the transition fro

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$2$

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(A) The energy of a state is given by $E_n = -\frac{13.6 Z^2}{n^2}$. Let $E_1 = -13.6 Z^2$.The most energetic photon corresponds to the transition from the $n^{th}$ state to the ground state $(n=1)$: $E_{max} = E_n - E_1 = -\frac{E_1}{n^2} - E_1 = -E_1 \left( \frac{1}{n^2} - 1 \right) = 52.224 \ eV$ ... $(i)$The least energetic photon corresponds to the transition between adjacent states ($n$ to $n-1$):$E_{min} = E_n - E_{n-1} = -\frac{E_1}{n^2} - \left( -\frac{E_1}{(n-1)^2} \right) = E_1 \left( \frac{1}{(n-1)^2} - \frac{1}{n^2} \right) = 1.224 \ eV$ ... $(ii)$Dividing $(i)$ by $(ii)$:$\frac{1 - 1/n^2}{1/(n-1)^2 - 1/n^2} = \frac{52.224}{1.224} = 42.666... \approx 42.67$Solving for $n$, we find $n=5$.Substituting $n=5$ into $(i)$:$-E_1 (1/25 - 1) = 52.224 \implies E_1 (24/25) = 52.224 \implies E_1 = 54.4 \ eV$.Since $E_1 = -13.6 Z^2 = -54.4$, we get $Z^2 = 4$, so $Z = 2$.

$A$. Speed of revolution of electron	$i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
$B$. Kinetic energy	$ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$C$. Total energy	$iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$D$. Frequency	$iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$

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