Chapter 7, Problem 7.109QP

Interpretation Introduction

Interpretation:

The longest wavelength associated with the Lyman series, the shortest wavelength associated with the Balmer series (in nanometers) and verification for not overlapping the spectral lines of the Lyman and Balmer series should be explained using the concept of Bohr’s theory. 

Concept Introduction:

The emission of radiation given by an energized hydrogen atom to the electron falling from a higher-energy orbit to a lower orbit give a quantum of energy in the form of light.  Based on electrostatic interaction and law of motion, Bohr derived the following equation.

${\text{E}}_{\text{n}}=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{n}}^{\text{2}}}\right)$

Where, $\text{n}$ gets an integer values such as $\text{n }=\text{ 1, 2, 3}$ and so on.  This is the energy of electron in ${\text{n}}^{\text{th}}$ orbital.

The electrons are excited thermally when the light is used by an object.  As a result, an emission spectrum comes.  Line spectra consist of light only at specific, discrete wavelengths.  In emission, the electron returns to a lower energy state from ${\text{n}}_{\text{f}}$ (the $\text{i and f}$ subscripts denote the initial and final energy states).  In most cases, the lower energy state corresponds to the ground state but it may be any energy state which is lower than the initial excited state.  The difference in the energies between the initial and final states is

$\text{ΔE }={\text{ E}}_{\text{f}}-{\text{ E}}_{\text{i}}$

This transition results in the photon’s emission with frequency $\text{v}$ and energy $\text{hv}$.  The following equation is resulted.

$\text{ΔE }=\text{ hν }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)$

When ${\text{n}}_{\text{i}}{\text{ > n}}_{\text{f}}$, a photon is emitted.  The term in parentheses is positive, making $\text{ΔE}$ negative.  As a result, energy is lost to the surroundings.  When ${\text{n}}_{\text{i}}{\text{ < n}}_{\text{f}}$, a photon is absorbed.  The term in parentheses is negative, so $\text{ΔE}$ is positive.  As a result, energy is absorbed from the surroundings.

To find: The longest wavelength associated with the Lyman series, the shortest wavelength associated with the Balmer series (in nanometers) and verification for not overlapping the spectral lines of the Lyman and Balmer series

Answer to Problem 7.109QP

The longest wavelength associated with the Lyman series is $\text{122 nm}$ whereas the shortest wavelength associated with the Balmer series is $\text{365 nm}$.  Hence, the spectral lines of the Lyman and Balmer series don’t overlap.

Explanation of Solution

The energy difference ($\text{ΔE}$) between ${\text{n}}_{\text{i}}={\text{ 2 and n}}_{\text{f}}=\text{ 1}$ associated with the Lyman series is calculated using the formula:

$\begin{array}{l}\text{ΔE }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)\\ \text{ΔE }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{1}}^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)\\ \text{ΔE }=-\text{1}{\text{.635 × 10}}^{-\text{18}}\text{ J}\end{array}$

Therefore, the energy difference ($\text{ΔE}$) between ${\text{n}}_{\text{i}}={\text{ 2 and n}}_{\text{f}}=\text{ 1}$ is $\text{1}{\text{.635 × 10}}^{-\text{18}}\text{ J}$ where the negative sign is ignored.  The negative sign indicates that this energy is associated with an emission process.  The longest wavelength has the smallest energy in it.  By Bohr’s theory,

$\begin{array}{l}\text{ΔE }=\text{ hν}\\ \text{ΔE }=\frac{\text{hc}}{\text{λ}}\because \text{ ν }=\frac{\text{c}}{\text{λ}}\\ \text{λ }=\frac{\text{hc}}{\text{ΔE}}\end{array}$

Planck’s constant, $\text{h}$ is $\text{6}{\text{.63 × 10}}^{-\text{34}}\text{ Js}$ and the speed of light, $\text{c}$ is $\text{3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}$.  This formula is used to find the wavelength for Lyman series.  Substitute the given values in the formula:

$\begin{array}{l}\text{λ }=\frac{\text{hc}}{\text{ΔE}}\\ \text{λ }=\frac{\text{(6}{\text{.63 × 10}}^{-\text{34}}\text{ Js)(3}{\text{.00 × 10}}^{\text{8}}\text{ m/s)}}{\text{1}{\text{.635 × 10}}^{-\text{18}}\text{ J}}\\ \text{λ }=\text{ 1}{\text{.22 × 10}}^{-\text{7}}\text{ m}\\ \text{λ }=\text{ 122 nm}\end{array}$

Therefore, the longest wavelength associated with the Lyman series is $\text{122 nm}$.  The energy difference ($\text{ΔE}$) between ${\text{n}}_{\text{i}}=\infty {\text{ and n}}_{\text{f}}=\text{ 2}$ associated with the Balmer series is calculated using the formula:

$\begin{array}{l}\text{ΔE }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J}\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)\\ \text{ΔE }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J}\left(\frac{\text{1}}{{\infty }^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)\\ \text{ΔE }=-\text{5}{\text{.450 × 10}}^{-\text{19}}\text{ J}\end{array}$

Therefore, the energy difference ($\text{ΔE}$) between ${\text{n}}_{\text{i}}=\infty {\text{ and n}}_{\text{f}}=\text{ 2}$ is $\text{5}{\text{.450 × 10}}^{-\text{19}}\text{ J}$ where the negative sign is ignored.  The negative sign indicates that this energy is associated with an emission process.  The shortest wavelength has the highest energy in it.  By Bohr’s theory,

$\begin{array}{l}\text{ΔE }=\text{ hν}\\ \text{ΔE }=\frac{\text{hc}}{\text{λ}}\because \text{ ν }=\frac{\text{c}}{\text{λ}}\\ \text{λ }=\frac{\text{hc}}{\text{ΔE}}\end{array}$

Planck’s constant, $\text{h}$ is $\text{6}{\text{.63 × 10}}^{-\text{34}}\text{ Js}$; the speed of light, $\text{c}$ is $\text{3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}$.  This formula is used to find the wavelength for Balmer series.  Substitute the given values in the formula:

$\begin{array}{l}\text{λ }=\frac{\text{hc}}{\text{ΔE}}\\ \text{λ }=\frac{\text{(6}{\text{.63 × 10}}^{-\text{34}}\text{ Js)(3}{\text{.00 × 10}}^{\text{8}}\text{ m/s)}}{\text{5}{\text{.450 × 10}}^{-\text{19}}\text{ J}}\\ \text{λ }=\text{ 3}{\text{.65 × 10}}^{-\text{7}}\text{ m}\\ \text{λ }=\text{ 365 nm}\end{array}$

Therefore, the shortest wavelength associated with the Balmer series is $\text{365 nm}$.  Hence, the spectral lines of the Lyman and Balmer series don’t overlap.

Conclusion

The longest wavelength associated with the Lyman series, the shortest wavelength associated with the Balmer series (in nanometers) and verification for not overlapping the spectral lines of the Lyman and Balmer series are explained using the concept of Bohr’s theory.