In a Hydrogen atom there is transition of electron from one orbit to another. Choose the correct option given below in which the emitted photon will have the highest frequency?

n = 2 to n = 1

The correct answer is Option (3) → n = 2 to n = 1

Energy of emitted photon: $E = h\nu = E_i - E_f$

Highest frequency corresponds to the largest energy difference. In hydrogen atom, energy levels:

$E_n = -13.6 \frac{1}{n^2} \ \text{eV}$

Calculate energy differences:

n=3 → 2: $E_3 - E_2 = -13.6/9 - (-13.6/4) = 1.89 \ \text{eV}$

n=6 → 2: $E_6 - E_2 = -13.6/36 - (-13.6/4) = 2.88 \ \text{eV}$

n=2 → 1: $E_2 - E_1 = -13.6/4 - (-13.6) = 10.2 \ \text{eV}$

n=4 → 2: $E_4 - E_2 = -13.6/16 - (-13.6/4) = 2.55 \ \text{eV}$

Largest energy difference = 10.2 eV → highest frequency.

Correct option: n = 2 to n = 1