An electron in a hydrogen atom (Bohr model) makes a transition from n = 2 state to n = 1 state. The frequency of the emitted photon is (take R = 1.0 × $$10^7$$​ $$\text{m}{^{-1}}$$​):

Correct option is A

​$$\text{Given:} \\\text{Initial state, } n_2 = 2 \\\text{Final state, } n_1 = 1 \\\text{The energy difference between the states is given by:} \\E_2 - E_1 = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \, \text{eV} \\\text{Step 1: Calculate the energy difference:} \\\text{Using } n_2 = 2 \text{ and } n_1 = 1: \\E_2 - E_1 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \, \text{eV} \\E_2 - E_1 = 13.6 \times \frac{3}{4} = 10.2 \, \text{eV} \\\text{Step 2: Convert energy to Joules:} \\\text{Now, convert the energy from eV to Joules. We know that:} \\1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \\E = 10.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J} = 1.632 \times 10^{-18} \, \text{J}$$

$$\text{Step 3: Calculate the frequency of the emitted photon:} \\\text{The frequency of the emitted photon is given by the formula:} \\\nu = \frac{E}{h} \\\text{Where:} \\E = 1.632 \times 10^{-18} \, \text{J} \\h = 6.626 \times 10^{-34} \, \text{} \ \text{} \\\text{Substituting the values:} \\\nu = \frac{1.632 \times 10^{-18}}{6.626 \times 10^{-34}} = 2.46 \times 10^{15} \, \text{Hz}$$​​​