Question

Calculate the frequency of the spectral line corresponding to the transition of an electron from $n_1=2$ and $n_2=4$ in a hydrogen atom.

• A. $6.172\times {10}^{14}{s}^{-1}$
• B. $3.086\times {10}^6{s}^{-1}$
• C. $6.172\times {10}^6{s}^{-1}$
• D. $3.086\times {10}^{14}{s}^{-1}$

Answer 1

The correct option is A $6.172\times {10}^{14}{s}^{-1}$

$\overline{v}=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})n_1=2n_2=4\overline{v}=\frac{1}{\lambda }$
R = Rydberg constant = 109678 $c{m}^{-1}$
Substituting the values in the equation,
$\overline{v}=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})c{m}^{-1}\frac{1}{\lambda }=20,564.625c{m}^{-1}\lambda =4.86\times {10}^{-5}cm\lambda =4.86\times {10}^{-7}m\text{frequency}\left(v\right)=\frac{c}{\lambda }v=\frac{3\times {10}^{8}m{s}^{-1}}{4.86\times {10}^{-7}m}=6.172\times {10}^{14}{s}^{-1}$