Chapter 7, Problem 7.68QE

Interpretation Introduction

Interpretation:

The wavelength $\left(\text{nm}\right)$ of the light absorbed by ${\text{He}}^{+}$ ions when they are excited from Bohr orbit with $n$ equals to 3 to $n$ equals to 4 has to be determined.

Concept Introduction:

The wave nature of any light can be described by its frequency, wavelength, and amplitude. The wavelength $\left(\text{λ, lambda}\right)$ of the light is defined as the distance between two successive peaks. Its SI unit is meter. The frequency $\left(v\right)$ is defined as the number of waves that can pass through the one point in 1 second. Its SI unit is ${\text{s}}^{-1}$. The amplitude $\left(A\right)$ of the light wave is maximum height of a wave.

The relation between frequency $\left(v\right)$ and wavelength $\left(\text{λ}\right)$ of the light is as follows:

  $v=\frac{c}{\text{λ}}$        (1)

Here, $c$ is the speed of light and its value is $3.00\times {10}^8\text{ m/s}$.

The relation between the energy $\left(E\right)$ of a photon and frequency $\left(v\right)$ is as follows:

  $E=hv$        (2)

Here, $h$ is known as Plank’s constant and its value is $\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}$.

The expression to calculate the energy of each wave function for any atomic species with one electron is as follows:

  $E_n=\frac{-2.18\times {10}^{-18}{\text{Z}}^2\text{ J}}{n^2}$        (3)

Here,

$Z$ is the nuclear charge or number of protons in nucleus of atomic species.

$n$ is the principal quantum number for Bohr orbit.

Answer to Problem 7.68QE

The wavelength of the light absorbed by ${\text{He}}^{+}$ ions when they are excited from Bohr orbit with $n$ equals to 3 to $n$ equals to 4 is $469\text{ nm}$.

Explanation of Solution

Substitute 3 for $n$ and 2 for $Z$ in equation (3).

  $\begin{array}{c}{E}_3=\frac{-2.18\times {10}^{-18}{\left(2\right)}^2\text{ J}}{{\left(3\right)}^2}\\ =-9.69\times {10}^{-19}\text{ J}\end{array}$

Substitute 4 for $n$ and 2 for $Z$ in equation (3).

  $\begin{array}{c}{E}_4=\frac{-2.18\times {10}^{-18}{\left(2\right)}^2\text{ J}}{{\left(4\right)}^2}\\ =-5.45\times {10}^{-19}\text{ J}\end{array}$

The expression to calculate energy absorbed in the excitation from $n$ equals to 3 to 4 is as follows:

  $\Delta E\left(4\to 1\right)=E_4-E_3$        (4)

Substitute $-5.45\times {10}^{-19}\text{ J}$ for $E_4$ and $-9.69\times {10}^{-19}\text{ J}$ for $E_3$ in equation (4).

  $\begin{array}{c}\Delta E\left(3\to 4\right)=-5.45\times {10}^{-19}\text{ J}-\left(-9.69\times {10}^{-19}\text{ J}\right)\\ =4.24\times {10}^{-19}\text{ J}\end{array}$

Therefore, the energy absorbed in the excitation is $4.24\times {10}^{-19}\text{ J}$.

Substitute $\frac{c}{\text{λ}}$ for $v$ in equation (2).

  $E=h\left(\frac{c}{\text{λ}}\right)$        (5)

Rearrange equation (5) to calculate the value of $\text{λ}$ as follows:

  $\text{λ}=\frac{hc}{E}$        (6)

Substitute $4.24\times {10}^{-19}\text{ J}$ for $E$, $\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}$ for $h$, and $3.00\times {10}^8\text{ m/s}$ for $c$ in equation (6).

  $\begin{array}{c}\text{λ}=\frac{\left(\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{ m/s}\right)}{4.24\times {10}^{-19}\text{ J}}\\ =\left(4.69\times {10}^{-7}\text{ m}\right)\left(\frac{\text{1 nm}}{{\text{10}}^{-9}\text{ m}}\right)\\ =469\text{ nm}\end{array}$