JEE Main 2023 — Atomic Physics Question with Solution

According to Bohr's atomic theory, whenever an electron jumps from one orbit to another, there is always an emission or absorption energy in terms of emission or absorption of photons.

The formula to calculate the energy $\left(∆E\right)$ of the emitted or absorbed photon when an electron jumps from a state with quantum number $n_1$ to another with quantum number $n_2$ is given by

$∆E= -Rch{Z}^2\left[\frac{1}{{n_i}^2}-\frac{1}{{n_f}^2}\right]....................\left(1\right)$

where, $R$ is Rydberg's constant, $h$ is Planck's constant, $c$ is the speed of light and $Z$ is the atomic number.

Substitute the values of the given parameters into equation (1) to calculate the required energy.

$\begin{array}{rcl}∆E& =& -13.6\times {\left(4\right)}^2\left[\frac{1}{4^2}-\frac{1}{2^2}\right] \mathrm{eV}\\ & =& -13.6\times 16\times \left[\frac{1}{16}-\frac{1}{4}\right] \mathrm{eV}\\ & =& 40.8 \mathrm{eV}\end{array}$