Chapter 7, Problem 7.106QP

A hydrogen atom in the ground stale absorbs a photon whose wavelength is 95.0 nm. The resulting excited atom then emits a photon of 1282 nm. What are the regions of the electromagnetic spectrum for the radiations involved in these transitions? What is the principal quantum number of the final state resulting from the emission from the excited atom?

Interpretation Introduction

Interpretation:

The principal quantum number of level from the emission of excited atom has to be calculated.

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.

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

Relation between frequency and wavelength is,

$\text{C}\text{=}\text{λν}$

C is the speed of light.

$\text{ν}$ is the frequency.

$\text{λ}$ is wavelength.

$\text{E}\text{=}\text{hν}$

h is Planck’s constant ( $\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$ ) which relates energy and frequency.

$\text{ν}$ is the frequency.

E is energy of light particle.

Answer to Problem 7.106QP

The principal quantum number of level from the emission of excited Hydrogen atom is three.

Explanation of Solution

To calculate: The principal quantum number of level from the emission of excited Hydrogen atom.

The level to which $\text{95}\text{.0}\text{nm}\text{photon}$(UV radiation) can excite the ground state hydrogen atom is calculated below.

The energy of the UV radiation is calculated as follows,

$\begin{array}{l}\text{E}\text{=}\text{hν}\text{=}\frac{\text{hc}}{\text{λ}}\\ \\ \text{=}\frac{\text{(6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s)(}\text{2}\text{.998}\text{×}{\text{10}}^{\text{8}}\text{m/s)}}{\text{95}\text{.0}\text{×}{\text{10}}^{\text{-9}}\text{m}}\\ \\ \text{=}2.091\text{×}{\text{10}}^{\text{-18}}\text{J}\end{array}$

Absorption occurs and the change in energy is positive.

$\begin{array}{l}\text{ΔE}\text{=}\text{hν}\text{=}{\text{E}}_{\text{1}}\text{-}{\text{E}}_{\text{i}}\\ \text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\text{-}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\\ \text{2}\text{.091}\text{×}{\text{10}}^{\text{-18}}\text{J}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\text{-}\frac{{\text{-R}}_{\text{H}}}{\text{1}}\\ \text{2}\text{.091}\text{×}{\text{10}}^{\text{-18}}\text{J}\text{=}{\text{-R}}_{\text{H}}\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\text{-}\text{1}\right)\\ \\ \frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\text{-}\text{1}\text{=}\frac{\text{ΔE}}{{\text{-R}}_{\text{H}}}\text{=}\frac{\text{2}\text{.091}\text{×}{\text{10}}^{\text{-18}}\text{J}}{\text{-2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J}}\text{=}\text{-0}\text{.9596}\\ \\ {\text{n}}_{\text{f}}\text{=}{\left[\frac{\text{1}}{\text{1}\text{-}\text{0}\text{.9596}}\right]}^{\text{0}\text{.5}}\text{=}\text{4}\text{.98}\end{array}$

The excited level of hydrogen atom is five.  This excited level by emitting $\text{1282}\text{nm}$ photon (IR radiation) goes to lower quantum state.

The energy of the IR radiation is calculated as follows,

$\begin{array}{l}\text{E}\text{=}\text{hν}\text{=}\frac{\text{hc}}{\text{λ}}\\ \\ \text{=}\frac{\text{(6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s)(}\text{2}\text{.998}\text{×}{\text{10}}^{\text{8}}\text{m/s)}}{\text{1282}\text{×}{\text{10}}^{\text{-9}}\text{m}}\\ \\ \text{=}\text{1}\text{.549}\text{×}{\text{10}}^{\text{-19}}\text{J}\end{array}$

Emission occurs and the change in energy is negative.

$\begin{array}{l}\text{ΔE}\text{=}\text{hν}\text{=}{\text{E}}_{\text{1}}\text{-}{\text{E}}_{\text{i}}\\ \text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\text{-}\frac{{\text{-R}}_{\text{H}}}{5^{\text{2}}}\\ -1.549\text{×}{\text{10}}^{\text{-19}}\text{J}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\text{-}\frac{{\text{-R}}_{\text{H}}}{5}\\ -1.549\text{×}{\text{10}}^{\text{-19}}\text{J}\text{=}{\text{-R}}_{\text{H}}\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\text{-}\frac{1}{5}\right)\\ \\ \frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\text{-}0.04\text{=}\frac{\text{ΔE}}{{\text{-R}}_{\text{H}}}\text{=}\frac{-1.549\text{×}{\text{10}}^{\text{-19}}\text{J}}{\text{-2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J}}\text{=}\text{-0}\text{.071087}\\ \\ {\text{n}}_{\text{f}}\text{=}{\left[\frac{\text{1}}{0.04+\text{0}\text{.071087}}\right]}^{\text{0}\text{.5}}\text{=}3.00\end{array}$

Conclusion

By using the formula for energy levels of electron in H-atom, the principal quantum number of level from the emission of excited atom was calculated.