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