Question

A stationary ${He}^{+}$ ion emitted a photon corresponding to a first line of the Lyman series. The photon liberated a photoelectrons from a stationary $H$ atom in ground state. What is the velocity of photoelectrons?

Answer 1

Energy of proton liberated by ${He}^{+}$

$H_{\alpha }$ line of Lyman series can be calculated as:$h_c{R}_H{Z}^2[\frac{1}{1^2}-\frac{1}{2^2}]$

$6.625\times {10}^{-27}\times 3\times 3\times {10}^8\times 19678\times 2^2\times \frac{3}{4}\Rightarrow 6.5\times {10}^{-11}erg$

$E_1=6.5\times {10}^{-11}erg$

$E_1=6.5\times {10}^{-11}ergofH+\frac{1}{2}m{u}^2$

$\Rightarrow 13.6\times 1.602\times {10}^{-21}+\frac{1}{2}m{u}^2$

$\frac{1}{2}m{u}^2=6.54\times {10}^{-11}-2.179\times {10}^{-11}$

$\frac{1}{2}m{u}^2=4.316\times {10}^{-11}erg$

$u^2=\frac{4.361\times {10}^{-11}\times 2}{9.108\times {10}^{-28}}$

$u=3.09\times {10}^{8}cm/s$