Question

The frequency of radiation emitted when the electron falls from$n=4$ to $n=1$ in a hydrogen atom will be: [ Given ionization energy of $H=2.18\times {10}^{-18}Jato{m}^{-1}$ and $h=6.625{10}^{-34}Js$ ]

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

Answer 1

The correct option is B $3.08\times {10}^{15}{s}^{-1}$

The transition energy for H-atom is given as:

$\Delta E=-R_H(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

$n_f=1$ and $n_i=4$

for ionization $n_f=\infty$ and $n_i=1$ and $\Delta E_{\infty }=2.18\times {10}^{-18}J/atom$

$\Delta E_{\infty }=-R_H(\frac{1}{{\infty }^2}-\frac{1}{1^2})$

$2.18\times {10}^{-18}J=R_H$

for $4\to 1$ transition

$\Delta E=-2.18\times {10}^{-18}(\frac{1}{1^2}-\frac{1}{4^2})$

$\Delta E=2.044\times {10}^{-18}J$

as $\nu =E/h$

therefore emission frequency is: $\nu =\frac{2.044\times {10}^{-18}}{6.63\times {10}^{-34}}$

$\nu =3.085\times {10}^{15}{s}^{-1}$

option B is correct