Calculate the wavelength of the radiation emitted when an electron in the hydrogen atom undergoes a transition from 4th energy level to the 2nd energy level

486.3nm

The correct answer is option 1. 486.3nm.

\(\text{For hydrogen atom, }E_n\text{ = –}\frac{\text{21.8 × }10^{-19}}{n^2}J/atom\)

\(\text{Energy emitted when the electron jumps from n=4 to n=2 will be given by}\)

\(\Delta \text{E = }E_4\text{ – }E_2\text{ = 21.8 × }10^{-19}(\frac{1}{2^2}\text{ – }\frac{1}{4^2})\text{ = 21.8 × }10^{-19}\text{ × }\frac{3}{16}\text{ = 4.0875 × }10^{-19}J\)

\(\text{The wavelength corresponding to this energy can be calculated using the expression, }\)

\(\text{E = h}\nu \text{= h}\frac{c}{\lambda}\text{       since, c = }\nu \lambda \)

\(\lambda \text{ = }\frac{hc}{E}\text{ = }\frac{(\text{6.626 ×}10^{-34}Js)(\text{3 × }10^8ms^{-1})}{\text{ = 4.0875 × }10^{-19}J}\text{ = 4.863 × }10^{-7}m\text{ = 486.3nm}\text{ = 4863}\overset{o}{A}\)

\(\text{It lies in the visible region.}\)