Question

A stationary $H{e}^{+}$ ion emitted a photon corresponding to the first line of the Lyman series. That photon liberated a photo electron from a stationary $H$ atom in ground state. What is the velocity of photo electron ?
$RH=109678c{m}^{-1}$

• A. $9.1\times 10^{8}cmse{c}^{-1}$
• B. $5.1\times 10^{8}cmse{c}^{-1}$
• C. $6.1\times 10^{8}cmse{c}^{-1}$
• D. $3.1\times 10^{8}cmse{c}^{-1}$

Answer 1

Answer: (D)

Energy of photon liberated from $H{e}^{+}$during emission of first line of Lyman series
$=hC\cdot R_H{Z}^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$
$=6.625\times 10^{-27}\times 3\times 10^{10}\times 109678\times 2^2\left[\frac{3}{4}\right]$
$=6.54\times 10^{-11}$ erg
This energy is used in liberating electron from $H$ atom from ground state, therefore,
$6.54\times 10^{-11}=E_1$ of $H+\frac{1}{2}m{u}^2$
$=13.6\times 1.602\times 10^{-12}+\frac{1}{2}m{u}^2$
$\frac{1}{2}m{u}^2=6.654\times 10^{-11}-2.179\times 10^{-11}$
$=4.361\times 10^{-11}erg$
$u^2=\frac{4.361\times 10^{-11}\times 2}{9.108\times 10^{-28}}$
$u=3.09\times 10^{8}cmse{c}^{-1}$