Chapter 8, Problem 8.118QP

Interpretation Introduction

Interpretation:

The kinetic energy of an electron emitted from Helium ion has to be calculated.

Concept introduction:

The kinetic energy of an electron emitted is calculated using given formula

Bohr formula is when the $n_f=\infty ,n_i=1andZ=2$

${\text{ΔE=-R}}_{\text{H}}\left[\frac{2^2}{{\infty }^2}-\frac{2^2}{1^2}\right]={\text{-R}}_{\text{H}}\left[\text{-4}\right]\text{=}\text{4}{\text{R}}_{\text{H}}$

Answer to Problem 8.118QP

The kinetic energy of an electron emitted from Helium ion ionized is $\text{4}\text{.5}\text{×}{\text{10}}^{\text{-18}}\text{J}$

Explanation of Solution

Bohr formula is

${\text{ΔE=-R}}_{\text{H}}\left[\frac{2^2}{{\infty }^2}-\frac{2^2}{1^2}\right]={\text{-R}}_{\text{H}}\left[\text{-4}\right]\text{=}\text{4}{\text{R}}_{\text{H}}$

$\frac{\text{4}\text{×}\text{2}{\text{.179×10}}^{\text{-18}}\text{J}}{\text{1}{\text{He}}^{\text{+}}\text{ion}}$

$\text{I}\text{.E=}\frac{\text{4}\text{×}\text{2}{\text{.179×10}}^{\text{-18}}\text{J}}{\text{1}{\text{He}}^{\text{+}}\text{ion}}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}{\text{He}}^{\text{+}}\text{ions}}{\text{1}{\text{He}}^{\text{+}}\text{ions}}\text{=}\frac{\text{5}\text{.2}\underset{\_}{\text{4}}{\text{87×10}}^{\text{6}}\text{J}}{\text{1}\text{mol}{\text{He}}^{\text{+}}}$

$\text{I}\text{.E}\text{.=5}{\text{.25×10}}^{\text{3}}\text{kJ/mol}{\text{He}}^{\text{+}}$

The electron is removed from H atom

The law of conservation of energy,

${\text{ E}}_{\text{photon}}\text{=}{\text{E}}_{\text{kinetic}}\text{+}\text{I}\text{.E}$

The energy of the 15 nm photon is calculated

$\begin{array}{l}{\text{ E}}_{\text{photon}}\text{=}\text{hυ}\text{=}\frac{\text{hc}}{\text{λ}}\text{=}\frac{\text{(6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\frac{\text{J}\text{.s}}{\text{photon}}\text{)(2}\text{.998×}\text{10}{}^{\text{8}}\text{m/s)}}{\text{15}\text{.0}\text{×}{\text{10}}^{\text{-9}}\text{m}}\\ \text{=}0.4\underset{\_}{5}2\text{×}{\text{10}}^{\text{-17}}\frac{\text{J}}{\text{photon}}\end{array}$

$\text{ 8}\text{.716×}{\text{10}}^{\text{-18}}\text{J}$

  $\begin{array}{l}{\text{ E}}_{\text{kinetic}}\text{=}{\text{E}}_{\text{photon}}\text{-}\text{I}\text{.E}\text{=}\text{1}\text{.3}\underset{\_}{\text{2}}\text{4}\text{×}\text{10}{}^{\text{-17}}\text{J}\text{-}\text{8}\text{.716×}\text{10}{}^{\text{-18}}\text{J}\text{=0}\text{.4}\underset{\_}{\text{5}}\text{2}\text{×}{\text{10}}^{\text{-17}}\text{J}\\ \text{=}\text{4}\text{.5}\text{×}{\text{10}}^{\text{-18}}\text{J}\end{array}$

Conclusion

The kinetic energy of an electron emitted from Helium ion ionized as found to be  $\text{4}\text{.5}\text{×}{\text{10}}^{\text{-18}}\text{J}$