Chapter 7, Problem 7.84QP

Interpretation Introduction

Interpretation:

The wavelength of emitted light when an electron from hydrogen atom transit from $\text{n}\text{=}4\text{to}\text{n}\text{=}2$ has to be calculated.  Also where the emission occurs in electromagnetic spectrum has to be explained.

Concept introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \text{Eis}\text{energy}\text{level}\text{.}\\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

The energy lost during a transition is calculated using Balmer’s formula

$\begin{array}{l}\text{ΔE}\text{=}{\text{E}}_{\text{f}}\text{-}\text{E}{}_{\text{i}}\\ \\ \text{=}\left(\frac{{\text{-R}}_{\text{H}}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\right)\text{-}\left(\frac{{\text{-R}}_{\text{H}}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)\text{=}{\text{-R}}_{\text{H}}\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)\end{array}$

Answer to Problem 7.84QP

The wavelength of emitted light when an electron from hydrogen atom transit from $\text{n}\text{=}4\text{to}\text{n}\text{=}2$ is $\text{54 nm}$.

The emission occurs in far UV region of electromagnetic spectrum

Explanation of Solution

To calculate: The wavelength of emitted light when an electron from hydrogen atom transit from $\text{n}\text{=}4\text{to}\text{n}\text{=}2$.

Energy levels are given as $\frac{{\text{-Z}}^{\text{2}}{\text{R}}_{\text{H}}}{{\text{n}}^{\text{2}}}$.

Atomic number of lithium is $\text{Z}\text{=}3$.

$\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}$

$\begin{array}{l}\text{ΔE}\text{=}\text{hν}\text{=}{\text{E}}_4\text{-}{\text{E}}_2\\ \text{=}\frac{{\text{-(3}}^{\text{2}}{\text{)R}}_{\text{H}}}{4^{\text{2}}}\text{-}\frac{{\text{-(3}}^{\text{2}}{\text{)R}}_{\text{H}}}{2^{\text{2}}}\\ \text{=}\frac{{\text{-9R}}_{\text{H}}}{16}\text{-}\frac{{\text{-9R}}_{\text{H}}}{4}\\ \text{=}\frac{{\text{108R}}_{\text{H}}}{64}\end{array}$

$\begin{array}{l}\text{ν}\text{=}\frac{{\text{108R}}_{\text{H}}}{\text{64h}}\text{=}\frac{108}{64}\text{×}\frac{\text{2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J}}{\text{6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.}\text{s}}\\ \\ \text{=}\text{5}\text{.5494}\text{×}{\text{10}}^{\text{15}}\text{/s}\end{array}$

The wavelength of emitted light is calculated as follows,

$\begin{array}{l}\text{C}\text{=}\text{λν}\\ \\ \text{λ}\text{=}\frac{\text{c}}{\text{ν}}\text{=}\frac{\text{2}\text{.998}\text{×}{\text{10}}^{\text{8}}\text{m/s}}{\text{5}\text{.5494}\text{×}{\text{10}}^{\text{15}}\text{/s}}\text{=}\text{5}\text{.402}\text{×}{\text{10}}^{\text{-8}}\text{m (54 nm)}\end{array}$

The emission occurs in far UV region of electromagnetic spectrum.

Conclusion

By using the formula for energy levels of electron in H-atom, the wavelength of emitted light when an electron from hydrogen atom transit from $\text{n}\text{=}4\text{to}\text{n}\text{=}2$ was calculated.