Question

The wave number of hydrogen atom in Lymen series is $82200\, cm^{-1}$. 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, $R=109677 \cong 109600\, cm ^{-1}$

and $\bar{v}=82200\, cm ^{-1}$

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 $\left(n_{1}\right)$