Chapter 7, Problem 7.34QP

Interpretation Introduction

Interpretation:

The value of ${\text{n}}_{\text{i}}$ in which the photon emitted that has a wavelength of $\text{434 nm}$ and the involved transition from an energy state of principal quantum number ${\text{n}}_{\text{i}}$ to the $\text{n }=\text{ 2}$ state of an electron in the hydrogen atom should be calculated 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.

The speed, wavelength and frequency of a wave are interrelated by $\text{c }=\text{ λν}$ where $\text{λ and ν}$ are mentioned in meters ($\text{m}$) and reciprocal seconds (${\text{s}}^{-1}$).    Hence, rearrange the equation for getting the wavelength is,

$\text{ν }=\frac{\text{c}}{\text{λ}}$

Substitute the frequency formula and rearrange,

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

Therefore, this formula is used to find the wavelength of the given photon in the emission line process.  For the absorption line process in which an electron is removed from the nucleus, the sign is changed as,

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

To find: Get the value of ${\text{n}}_{\text{i}}$ in which the photon emitted that has a wavelength of $\text{434 nm}$ and the involved transition from an energy state of principal quantum number ${\text{n}}_{\text{i}}$ to the $\text{n }=\text{ 2}$ state of an electron in the hydrogen atom

Answer to Problem 7.34QP

The value of ${\text{n}}_{\text{i}}$ in which the photon emitted that has a wavelength of $\text{434 nm}$ and the involved transition from an energy state of principal quantum number ${\text{n}}_{\text{i}}$ to the $\text{n }=\text{ 2}$ state of an electron in the hydrogen atom is 5.

Explanation of Solution

An electron in the hydrogen atom makes a transition from an energy state of principal quantum number  ${\text{n}}_{\text{i}}$ to the $\text{n }=\text{ 2}$.  The photon emitted has a wavelength of $\text{434 nm}$.  The formula used to find the value of ${\text{n}}_{\text{i}}$ is 

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

Here, Planck’s constant, $\text{h}$ is $\text{6}{\text{.63 × 10}}^{-\text{34}}\text{ Js}$; the speed of light, $\text{c}$ is $\text{3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}$; $\text{λ }=\text{ 434 nm}$ and ${\text{n}}_{\text{f}}=\text{ 2}$.  To find the value of ${\text{n}}_{\text{i}}$, $\text{ΔE}$ should be calculated first. 

$\begin{array}{l}\text{ΔE = }\frac{\text{(6}{\text{.63 × 10}}^{-\text{34}}\text{ Js)(3}{\text{.00 × 10}}^{\text{8}}\text{ m/s)}}{{\text{434 × 10}}^{-\text{9}}\text{ m}}\\ \text{ΔE = 4}{\text{.58 × 10}}^{-\text{19}}\text{ J}\end{array}$

Hence, the photon energy change is $-\text{4}{\text{.58 × 10}}^{-\text{19}}\text{ J}$ which should be negative because this is an emission process.  Substitute the given values in the formula:

$\begin{array}{l}\text{ΔE }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J}\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)\\ -\text{4}{\text{.58 × 10}}^{-\text{19}}\text{ J }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J}\left(\frac{\text{1}}{{\text{2}}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)\\ \frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}=\left(\frac{-\text{4}{\text{.58 × 10}}^{-\text{19}}\text{ J}}{\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J}}\right)\text{ + }\frac{\text{1}}{{\text{2}}^{\text{2}}}\\ \frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}=-\text{0}\text{.210 + 0}\text{.250}\\ \frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}=\text{ 0}\text{.040}\\ {\text{n}}_{\text{i}}=\frac{\text{1}}{\sqrt{\text{0}\text{.040}}}\\ {\text{n}}_{\text{i}}=\text{ 5}\end{array}$

Therefore, the value of ${\text{n}}_{\text{i}}$ in which the photon emitted that has a wavelength of $\text{434 nm}$ and the involved transition from an energy state of principal quantum number ${\text{n}}_{\text{i}}$ to the $\text{n }=\text{ 2}$ state of an electron in the hydrogen atom is 5.

Conclusion

The value of ${\text{n}}_{\text{i}}$ in which the photon emitted that has a wavelength of $\text{434 nm}$ and the involved transition from an energy state of principal quantum number ${\text{n}}_{\text{i}}$ to the $\text{n }=\text{ 2}$ state of an electron in the hydrogen atom should be calculated using the concept of Bohr’s theory.