Question

The frequency of radiation emitted when the electron falls from $ n=4 $ to $ n=1 $ in a hydrogen atom will be (Given ionisation energy of $ H= $ $ 2.18\,{{10}^{-18}}J\,ato{{m}^{-1}} $ and $ h=6.625\times {{10}^{-34}}Js $ )

• A. $ 1.54\times {{10}^{15}}{{s}^{-1}} $
• B. $ 1.03\times {{10}^{15}}{{s}^{-1}} $
• C. $ 3.08\times {{10}^{15}}{{s}^{-1}} $
• D. $ 2.00\times {{10}^{15}}{{s}^{-1}} $

Answer 1

Answer: (C)

Ionisation energy of $H$
$=2.18 \times 10^{-18} J \text { atom }^{-1}$
$ \therefore E_{1} $ (Energy of Ist orbit of $H$ -atom )
$=-2.18 \times 10^{-18} J - atom ^{-1} $
$ \therefore E_{n} =\frac{-2.18 \times 10^{-18}}{n^{2}} J - atom ^{-1}$
$=1 $ for $ H $-atom
$ \Delta E =E_{4}-E_{1} $
$=\frac{-2.18 \times 10^{-18}}{4^{2}}-\frac{-2.18 \times 10^{-18}}{1^{2}} $
$=-2.18 \times 10^{-18} \times\left[\frac{1}{4^{2}}-\frac{1}{1^{2}}\right]$
$\Delta E=h v=-2.18 \times 10^{-18} \times-\frac{15}{16}$
$=+2.0437 \times 10^{-18} \,J atom ^{-1}$
$\therefore v =\frac{\Delta E}{h}=\frac{2.0437 \times 10^{-18} J atom ^{-1}}{6.625 \times 10^{-34} J s }$
$=3.084 \times 10^{15} s ^{-1} atom ^{-1}$