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
$E = 13.6\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = 13.6 \times \dfrac{3}{16} = 2.55$ eV.
Solution by Sreeraj P, M.Sc Physics