Condition for emission vs absorption

In a hydrogen atom, a photon is emitted only when the electron goes from a higher energy level to a lower energy level.

So among the given transitions:

• A: $n=1 \to 2$ → absorption
• B: $n=2 \to 6$ → absorption
• C: $n=6 \to 2$ → emission
• D: $n=2 \to 1$ → emission

Hence, only C and D are emission transitions.

Frequency of emitted photon

The photon frequency is related to energy difference by $h\nu = \Delta E$ where for hydrogen, $E_n = -\frac{13.6}{n^2}\text{ eV}$

Greater the energy difference, greater the frequency.

Calculate energy difference for option C: $n=6 \to 2$

$\Delta E_C = 13.6\left(\frac{1}{2^2}-\frac{1}{6^2}\right)$ $=13.6\left(\frac{1}{4}-\frac{1}{36}\right)$ $=13.6\left(\frac{9-1}{36}\right)$ $=13.6\cdot \frac{8}{36}$ $=13.6\cdot \frac{2}{9} \approx 3.02\text{ eV}$

Calculate energy difference for option D: $n=2 \to 1$

$\Delta E_D = 13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$ $=13.6\left(1-\frac{1}{4}\right)$ $=13.6\cdot \frac{3}{4}$ $=10.2\text{ eV}$

Compare the two emission transitions

$\Delta E_D > \Delta E_C$ Therefore, $\nu_D > \nu_C$

Final answer

The transition that emits the photon of highest frequency is: $\boxed{n=2 \to n=1}$ i.e. Option D.