Question

Calculate the frequency of the spectral line corresponding to $n_1=2$ and $n_2=4$ of hydrogen atom. Also find the series this line belongs?

• A. $6.172\times {10}^{14}{\text{s}}^{-1}$, Balmer series
• B. $6.172\times {10}^{14}{\text{s}}^{-1}$, Brackett Series
• C. $5.182\times {10}^{14}{\text{s}}^{-1}$, Balmer Series
• D. $8.277\times {10}^{16}{\text{s}}^{-1}$, Lyman Series

Answer 1

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

We know that,
$\text{wave~number},\stackrel{-}{v}=\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
Here, $n_1=2,n_2=4,R\left(\text{Rydberg constant}\right)=109677{\text{cm}}^{-1}$
Substituting the values in the equation,
$\frac{1}{\lambda }=109677(\frac{1}{2^2}-\frac{1}{4^2}){\text{cm}}^{-1}=20564.43{\text{cm}}^{-1}$

$\Rightarrow \lambda =4.86\times {10}^{-5}\text{cm}=4.86\times {10}^{-7}\text{m}$

$\because v=\frac{c}{\lambda }=\frac{3\times {10}^8{\text{m s}}^{-1}}{4.86\times {10}^{-7}\text{m}}=6.172\times {10}^{14}{\text{s}}^{-1}$

Since $n_1=2$, so it belongs to the Balmer series.