SOLUTION —
We know that nave number
$\bar{v}=\frac{1}{\lambda}=R_{H}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right) ; \text { Given } n_{i}=1, n_{f}=3, R_{H}=109737 cm ^{-1}$
$\therefore \quad \frac{1}{\dot{i}}=109737\left(1-\frac{1}{3^{2}}\right)=\frac{109737 \times 8}{9} cm ^{-1}$
Frequency $(v)=\frac{c}{\lambda}=3.0 \times 10^{10} cm / sec . \times \frac{109737 \times 8}{9} cm ^{-1}$
$=\frac{3 \cdot 0 \times 1 \cdot 09737 \times 8}{9} \times 10^{15} \text { per sec. }= 2 . 9 2 6 \times 10^{15} H z .$
