Question

The ratio of energies of photons produced due to transition of an electron in hydrogen atom from second energy level to first energy level and fifth energy level to second energy level is

• A. 2:1
• B. 1:4
• C. 3:2
• D. 25:7

Hint:

Energy transition in hydrogen: $E \propto \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.

Answer 1

The Correct Option is D

The energy of a photon emitted during a transition from energy level $n_i$ to $n_f$ in a hydrogen atom is given by: $$ E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{eV} $$ For the transition from $n=2$ to $n=1$, $E_1 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \times \frac{3}{4} = 10.2 \, \text{eV}$ For the transition from $n=5$ to $n=2$, $E_2 = 13.6 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = 13.6 \times \frac{21}{100} = 2.856 \, \text{eV}$ The ratio of the energies is: $$ \frac{E_1}{E_2} = \frac{10.2}{2.856} = \frac{10.2}{2.856} \approx \frac{10.2}{2.86} \approx 3.57 \approx \frac{25}{7} $$