Q 12-12-025JEE MainEasy
Find the energy, in eV, of the photon emitted when the electron in a hydrogen atom falls from $n = 4$ to $n = 2$. (Give the answer to two decimal places.)
Numerical value type. Enter your answer.
Answer: 2.55
Concept. The photon carries the energy difference: $E = 13.6\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ eV.
Step 1. Substitute. $n_i = 4$, $n_f = 2$:
$$E = 13.6\left(\frac14 - \frac{1}{16}\right) = 13.6 \times \frac{3}{16} = 2.55\ \text{eV}$$
Answer. $2.55$.
Solution by Sreeraj P, M.Sc Physics