Question

Calculate the wavelength and energy of radiation emitted for the electronic transition from infinity to stationary state of the hydrogen.

• A. 2.178 $\times$ 10${}^{18}$ J
• B. 3.142 $\times$ 10${}^{18}$ J
• C. 1.021 $\times$ 10${}^{18}$ J
• D. 0.536 $\times$ 10${}^{18}$ J

Answer 1

The correct option is A 2.178 $\times$ 10${}^{18}$ J

$\frac{1}{\lambda }$ = $R$ [$\frac{1}{n_1^2}$ - $\frac{1}{n_2^2}$]
= $109678$ cm${}^{-1}$ [$\frac{1}{1^2}$ - $\frac{1}{{\infty }^2}$]
= $109678$ cm ${}^{-1}$
$\lambda$ = $\frac{1}{109678c{m}^{-1}}$ = 9.118 $\times$ 10${}^{-6}$ cm
E = $\frac{hc}{\lambda }$ = $\frac{6.62\times {10}^{-34}\times 3}{9.118\times {10}^{-8}}$ = 2.178 $\times$ 10${}^{18}$ J