Question

Find the wavelength of the photon emitted during the transition from energy level n = 6 to the n = 3 in a hydrogen atom.

• A. 1099 $\stackrel{o}{A}$
• B. 1099 nm
• C. 1310 nm
• D. 1310 $\stackrel{o}{A}$

Answer 1

The correct option is B 1099 nm

We know from Rydberg equation,
$\frac{1}{\lambda }=R_H\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$
Here, $n_1=3$
$n_2=6and{R}_H=109878c{m}^{-1}$
$\therefore \frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{3^2}-\frac{1}{6^2})$
$=0.091\times {10}^7{m}^{-1}$
$\Rightarrow \lambda =10.99\times {10}^{-7}m=1099nm$