JEE Physics 2023 Question with Solution

Given: A hydrogen-like atom with $Z = 4$ undergoes a transition from $n_i = 4$ to $n_f = 2$.

Find: The energy released during the transition and the correct option.

For a hydrogen-like atom, the energy of the electron in the $n^{\text{th}}$ orbit is

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

So, the emitted energy is

$$\Delta E = E_{n_i} - E_{n_f} = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \, \text{eV}$$

Substituting $Z = 4$, $n_i = 4$, and $n_f = 2$,

$$\Delta E = 13.6 \times (4^2) \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \, \text{eV}$$ $$\Delta E = 13.6 \times 16 \left( \frac{1}{4} - \frac{1}{16} \right) \, \text{eV}$$ $$\Delta E = 13.6 \times 16 \left( \frac{4 - 1}{16} \right) \, \text{eV}$$ $$\Delta E = 13.6 \times 16 \times \frac{3}{16} \, \text{eV}$$ $$\Delta E = 13.6 \times 3 = 40.8 \, \text{eV}$$

Therefore, the energy released is $40.8 \, \text{eV}$.

The solution working gives $40.8 \, \text{eV}$, which matches Option D. The solution labels the option as A and also says Option 1, which is inconsistent with the calculation. Hence, the defensible correct option from the working is D.