Bohr model energy levels

For a hydrogen-like atom, the energy of the electron in the $n^{\text{th}}$ orbit is $E_n=-\frac{13.6}{n^2}\,\text{eV}$

Frequency of emitted radiation

When an electron jumps from a higher level to a lower level, the emitted photon has energy $h\nu = \Delta E = E_{\text{final}}-E_{\text{initial}}$ Since frequency $\nu$ is directly proportional to the magnitude of energy difference, the transition with the largest energy gap will have the maximum frequency.

Calculate energy differences for each option

Option A: $n=2 \to n=1$

$\Delta E = 13.6\left(1-\frac{1}{2^2}\right)$ $=13.6\left(1-\frac14\right)=13.6\cdot \frac34=10.2\,\text{eV}$

Option B: $n=3 \to n=2$

$\Delta E = 13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ $=13.6\left(\frac14-\frac19\right)=13.6\cdot \frac{5}{36}\approx 1.89\,\text{eV}$

Option C: $n=4 \to n=3$

$\Delta E = 13.6\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$ $=13.6\left(\frac19-\frac{1}{16}\right)=13.6\cdot \frac{7}{144}\approx 0.66\,\text{eV}$

Option D: $n=5 \to n=4$

$\Delta E = 13.6\left(\frac{1}{4^2}-\frac{1}{5^2}\right)$ $=13.6\left(\frac{1}{16}-\frac{1}{25}\right)=13.6\cdot \frac{9}{400}\approx 0.306\,\text{eV}$

Compare the energy gaps

$10.2 > 1.89 > 0.66 > 0.306$

Therefore, the maximum frequency corresponds to the transition: $n=2 \to n=1$

Correct option

Option A is correct.