Question

Calculate the wavelength and energy of radiation emitted for the electronic transition from infinite to stationary state of hydrogen atom. (Given, $R=1.09678\times {10}^7{m}^{-1}$, $h=6.6256\times {10}^{-34}J-s$ and $c=2.9979\times {10}^{8}m{s}^{-1}$).

Answer 1

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$n_1=1$ and $n_2=\infty$
$\frac{1}{\lambda }=R[\frac{1}{1^2}-\frac{1}{{\left(\infty \right)}^2}]=R$
or $\lambda =\frac{1}{R}=\frac{1}{1.09678\times {10}^7}=9.11\times {10}^{-8}m$
We know that,
$E=hv=h\cdot \frac{c}{\lambda }=6.6256\times {10}^{-34}\times \frac{2.9979\times {10}^8}{9.11\times {10}^{-8}}$
$=2.17\times {10}^{-18}J$.