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 $H=$ $2.18{10}^{-18}Jato{m}^{-1}$ 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 )
$=-2.18\times 10^{-18}J-ato{m}^{-1}$
$\therefore E_n=\frac{-2.18\times 10^{-18}}{n^2}J-ato{m}^{-1}$
$=1$ for $H$-atom
$\Delta E=E_4-E_1$
$=\frac{-2.18\times 10^{-18}}{4^2}-\frac{-2.18\times 10^{-18}}{1^2}$
$=-2.18\times 10^{-18}\times \left[\frac{1}{4^2}-\frac{1}{1^2}\right]$
$\Delta E=hv=-2.18\times 10^{-18}\times -\frac{15}{16}$
$=+2.0437\times 10^{-18}Jato{m}^{-1}$
$\therefore v=\frac{\Delta E}{h}=\frac{2.0437\times 10^{-18}Jato{m}^{-1}}{6.625\times 10^{-34}Js}$
$=3.084\times 10^{15}{s}^{-1}ato{m}^{-1}$