Out of the following transitions in hydrogen atom, identify the transition which emits photons of highest frequency.

The photon frequency is proportional to the transition energy difference as given by $E = h\nu$.

Hydrogen energy levels follow $E_n = -\frac{13.6}{n^2}$ eV. Only transitions with $n_i > n_f$ result in photon emission.

Option A ($n=1 \rightarrow 2$) and C ($n=2 \rightarrow 6$) are absorption transitions—no emission occurs. For emission transition B ($n=2 \rightarrow 1$):
$\Delta E_B = E_2 - E_1 = 13.6 \left(1 - \frac{1}{4}\right) = 10.2$ eV

For D ($n=6 \rightarrow 2$):
$\Delta E_D = E_6 - E_2 = 13.6 \left(\frac{1}{4} - \frac{1}{36}\right) = 13.6 \cdot \frac{2}{9} \approx 3.02$ eV

Since $\Delta E_B > \Delta E_D$, transition B yields the highest frequency photon.