Question

The wave number of hydrogen atom in Lymen series is $82200c{m}^{-1}$. The electron goes from

• A. $n_3\to n_2$
• B. $n_2\to n_1$
• C. $n_4\to n_3$
• D. None of these

Answer 1

Answer: (B)

According to Rydberg formula ;

$\bar{v}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Here, $R=109677\cong 109600c{m}^{-1}$

and $\bar{v}=82200c{m}^{-1}$

thus, $\frac{82200}{109600}=\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right)$

or $\frac{3}{4}=\left(1-\frac{1}{n_2^2}\right)$

or $\frac{1}{n_2^2}=1-\frac{3}{4}=\frac{1}{4}$

or $n_2=2$

$\therefore$ The electron jumps from second orbit $\left(n_2\right)$ to ground state $\left(n_1\right)$