Atoms and Nuclei Part 1

Let $n_i$ and $n_f$ be the initial and final orbit numbers. For a hydrogen atom, the wave number is given by $$\frac{1}{\lambda}=R\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]$$ where $R=1.097\times10^{7}\,\text{m}^{-1}$ is the Rydberg constant. Since frequency is $\nu=\frac{c}{\lambda}$, we get $\nu\propto\frac{1}{\lambda}$, so the transition with the largest value of $\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]$ will have the maximum frequency. Checking each option, (a) $n=4$ to $n=3$: $\frac{1}{3^{2}}-\frac{1}{4^{2}}=\frac{1}{9}-\frac{1}{16}\approx0.05$ $\dots$ (i) (b) $n=2$ to $n=1$: $\frac{1}{1^{2}}-\frac{1}{2^{2}}=1-\frac{1}{4}=0.75$ $\dots$ (ii) (c) $n=5$ to $n=4$: $\frac{1}{4^{2}}-\frac{1}{5^{2}}=\frac{1}{16}-\frac{1}{25}\approx0.0225$ $\dots$ (iii) (d) $n=3$ to $n=2$: $\frac{1}{2^{2}}-\frac{1}{3^{2}}=\frac{1}{4}-\frac{1}{9}\approx0.14$ $\dots$ (iv) Option (b) has the highest value, i.e. the smallest wavelength and hence the highest frequency. Equivalently, using $\Delta E=13.6\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\text{eV}$ and $\Delta E=h\nu$, the energy gap is largest for $n=2$ to $n=1$ ($10.2\,\text{eV}$), so this transition emits the most energetic photon. $\therefore$ The frequency will be maximum for the transition $n=2$ to $n=1$.