Question

The wave number of the first balmer line of $L{i}^{2+}$ ions is 1,36,800 $c{m}^{-1}$. The wavelenght of the first line of Balmer series of hydrogen atom is (in $c{m}^{-1}$)

• A. 68,400
• B. 15,200
• C. 76,000
• D. 30,800

Answer 1

The correct option is B

15,200

Atomic number of $L{i}^{2+}$ is +3

$\frac{1}{\lambda }$=$R_n$$Z^2$($\frac{1}{n_1^2}$-$\frac{1}{n_2^2}$)
1,36,800=$R_n$$\times$9($\frac{1}{2^2}$-$\frac{1}{3^2}$)

$\frac{1}{\lambda }$ of H-atom =$R_n$($\frac{1}{2^2}$-$\frac{1}{3^2}$)=$\frac{1,36,800}{9}$

=15200 $c{m}^{-1}$