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

$\begin{array}{1 1} (a)\;2.186 \times 10^{-18}\;J \\ (b)\;3.186 \times 10^{-18}\;J \\ (c)\;5.186 \times 10^{-18}\;J\\(d)\;1.186 \times 10^{-18}\;J \end{array}$

$\large\frac{1}{\lambda}$$=109678 \bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg] \qquad= 109678 \bigg[\large\frac{1}{(1)^2}-\frac{1}{\infty}\bigg] \lambda= \large\frac{1}{109678}$$cm=9.1 \times 10^{-6}\;cm$
$\qquad= 9.1 \times 10^{-8}m =91\;nm$
$E= \large\frac{hc}{\lambda}=\frac{6.63 \times 10^{-34}\;J.s \times 3 \times 10^8\;ms^{-1}}{9.1 \times 10^{-8}\;m}$
$E= 2.186 \times 10^{-18}\;J$