Question

Calculate the energy of the photon emitted during the transition from energy levels n = 4 to n = 2 in a hydrogen atom.

• A. $4.09\times {10}^{-19}J$
• B. $2.42\times {10}^{-19}J$
• C. $5.42\times {10}^{-19}J$
• D. $3.42\times {10}^{-19}J$

Answer 1

The correct option is A $4.09\times {10}^{-19}J$

We know,
$\frac{1}{\lambda }=R_H\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$
Here, $n_1=2,n_2=4$
$\therefore \frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{2^2}-\frac{1}{4^2})$
$=0.206\times {10}^7{m}^{-1}$
$\Rightarrow \lambda =4.85\times {10}^{-7}m$
We also know,
$E=h\nu =\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{4.85\times {10}^{-7}}J=4.09\times {10}^{-19}J$
So, the energy of the emitted photon is $4.09\times {10}^{-19}J$.