Question

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

• A. $u=3.09\times {10}^{10}cmse{c}^{-1}$
• B. $u=3.09\times {10}^{8}mse{c}^{-1}$
• C. $u=3.09\times {10}^{9}cmse{c}^{-1}$
• D. $u=3.09\times {10}^{8}cmse{c}^{-1}$

Answer 1

Answer: (D)

$u=3.09\times {10}^{8}cmse{c}^{-1}$

Photon energy liberated from $H{e}^{+}=\left(hc/\lambda \right)$

In I${}^{st}$ line of Lyman series $=hc{R}_H\times Z^2\left[(\frac{1}{1^2}-\frac{1}{2^2})\right]$

$=6.626\times {10}^{-27}\times 3.0\times {10}^{10}\times 109678\times 2^2\times \left[3/4\right]$

$=6.54\times {10}^{-11}erg$

This energy is used in liberating electron from $H$-atom from the ground state and therefore

$6.54\times {10}^{-11}=E_1$ of $H+KE$ of electron given out [$E_H=2.18\times {10}^{-11}erg]$

$\therefore \frac{1}{2}m{u}^2=4.362\times {10}^{-11}$

or $u^2=\frac{4.362\times {10}^{-11}\times 2}{9.108\times {10}^{-28}}$

$\therefore u=3.09\times {10}^{8}cmse{c}^{-1}$