Question

A hydrogen-like atom is in a higher energy level of quantum number $6$. The excited atom make a transition to first excited state by emitting photons of total energy $27.2eV$. The atom from the same excited state make a transition to the second excited state by successively emitting two photons. If the energy of one photon is $4.25eV$, find the energy of other photon.

• A. $5.25eV$
• B. $6.25eV$
• C. $6.95eV$
• D. $7.80eV$

Answer 1

The correct option is A $5.25eV$

Total energy liberated during transition of $e^{-}$ from $n^{th}$ shell to first excited state

i.e, $2^{nd}$ shell $=10.20+17.0=2720eV$

$=27.20\times 1.602\times {10}^{-12}erg$

$\frac{hc}{\lambda }=R_H\times Z^2{h}_c[\frac{1}{z^2}-\frac{1}{n^2}]$

$27.20\times 1.602\times {10}^{-12}=R_H\times Z^2{h}_c[\frac{1}{z^2}-\frac{1}{n^2}]⟶\left(1\right)$

i.e. $3^{rd}$ shell $=4.25+5.95=10.20eV$

$=10.20\times 1.602\times {10}^{-12}erg$

$\therefore$ $10.20\times 1.602\times {10}^{-12}=R_H\times Z^2{h}_c[\frac{1}{3^2}-\frac{1}{n^2}]$ $⟶\left(2\right)$

We get $n=5.25eV$