Question

Calculate the frequency of the radiation emitted when an electron jumps from n=3 to n=2 in a hydrogen atom is:

• A. $4.57\times {10}^{14}{s}^{-1}$
• B. $4.57\times {10}^{16}{s}^{-1}$
• C. $5.01\times {10}^{14}{s}^{-1}$
• D. $5.01\times {10}^{16}{s}^{-1}$

Answer 1

Answer: (A)

$4.57\times {10}^{14}{s}^{-1}$

Wavelength is represented by $\lambda$ and is given by following expression
$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\frac{1}{\lambda }=\left(109677{cm}^{-1}\right)[\frac{1}{2^2}-\frac{1}{3^2}]$
$\Rightarrow \frac{1}{\lambda }=1523291.67m^{-1}$
Frequency $\nu$ is given by
$\nu =\frac{c}{\lambda }=4.57\times {10}^{14}{s}^{-1}$