Chapter 7, Problem 7.40QE

Interpretation Introduction

Interpretation:

The wavelength, initial quantum number and the final quantum number in the visible spectrum of hydrogen have to be labeled.

Concept Introduction:

The wave nature of any light can be described by its frequency, wavelength, and amplitude. The wavelength $\left(\lambda \text{, lambda}\right)$ 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 number of waves that can pass through 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 energy of Bohr orbits of hydrogen is dependent on the principal quantum number $\left(n\right)$. The expression to calculate energy of Bohr orbit is as follows:

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

Answer to Problem 7.40QE

The labeled visible spectrum of hydrogen is as follows:

Explanation of Solution

The visible spectrum of hydrogen gives four lines that represent four transitions of the hydrogen as follows:

(1) Transition 1 is from $n$ equals to 3 to 2.

(2) Transition 1 is from $n$ equals to 4 to 2.

(3) Transition 1 is from $n$ equals to 5 to 2.

(4) Transition 1 is from $n$ equals to 6 to 2.

The expression to calculate the energy difference between initial Bohr orbit $n_1$ and final Bohr orbit $n_2$ is as follows:

  $\Delta E\left(n_1\to n_2\right)=E_{n_2}-E_{n_1}$        (4)

Substitute $\frac{-2.18\times {10}^{-18}\text{ J}}{n_2^2}$ for $E_{n_2}$ and $\frac{-2.18\times {10}^{-18}\text{ J}}{n_1^2}$ for $E_{n_1}$ in equation (4).

  $\begin{array}{c}\Delta E\left(n_1\to n_2\right)=\frac{-2.18\times {10}^{-18}\text{ J}}{n_2^2}-\left(\frac{-2.18\times {10}^{-18}\text{ J}}{n_1^2}\right)\\ =-2.18\times {10}^{-18}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)\text{ J}\end{array}$

Hence, the expression to calculate energy difference between initial Bohr orbit $n_1$ and final Bohr orbit $n_2$ is,

  $\Delta E\left(n_1\to n_2\right)=-2.18\times {10}^{-18}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)\text{ J}$        (5)

Here,

$n_1$ is an initial Bohr orbit.

$n_2$ is final Bohr orbit.

For transition (1):

Substitute 3 for $n_1$ and 2 for $n_2$ in equation (5).

  $\begin{array}{c}\Delta E\left(n_1\to n_2\right)=-2.18\times {10}^{-18}\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2}\right)\text{ J}\\ =-3.03\times {10}^{-19}\text{ J}\end{array}$

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

  $E=h\left(\frac{c}{\text{λ}}\right)$        (6)

Rearrange equation (6) to calculate the value of $\text{λ}$ as follows:

  $\text{λ}=\frac{hc}{E}$        (7)

Substitute $3.03\times {10}^{-19}\text{ J}$ 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 (7).

  $\begin{array}{c}\text{λ}=\frac{\left(\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{ m/s}\right)}{3.03\times {10}^{-19}\text{ J}}\\ =\left(6.56\times {10}^{-7}\text{ m}\right)\left(\frac{\text{1 nm}}{{\text{10}}^{-9}\text{ m}}\right)\\ =656\text{ nm}\end{array}$

For transition (2):

Substitute 4 for $n_1$ and 2 for $n_2$ in equation (5).

  $\begin{array}{c}\Delta E\left(n_1\to n_2\right)=-2.18\times {10}^{-18}\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(4\right)}^2}\right)\text{ J}\\ =-4.09\times {10}^{-19}\text{ J}\end{array}$

Substitute $4.09\times {10}^{-19}\text{ J}$ 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 (7).

  $\begin{array}{c}\text{λ}=\frac{\left(\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{ m/s}\right)}{4.09\times {10}^{-19}\text{ J}}\\ =\left(4.86\times {10}^{-7}\text{ m}\right)\left(\frac{\text{1 nm}}{{\text{10}}^{-9}\text{ m}}\right)\\ =486\text{ nm}\end{array}$

For transition (3):

Substitute 5 for $n_1$ and 2 for $n_2$ in equation (5).

  $\begin{array}{c}\Delta E\left(n_1\to n_2\right)=-2.18\times {10}^{-18}\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(5\right)}^2}\right)\text{ J}\\ =-4.58\times {10}^{-19}\text{ J}\end{array}$

Substitute $4.58\times {10}^{-19}\text{ J}$ 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 (7).

  $\begin{array}{c}\text{λ}=\frac{\left(\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{ m/s}\right)}{4.58\times {10}^{-19}\text{ J}}\\ =\left(4.34\times {10}^{-7}\text{ m}\right)\left(\frac{\text{1 nm}}{{\text{10}}^{-9}\text{ m}}\right)\\ =434\text{ nm}\end{array}$

For transition (4):

Substitute 6 for $n_1$ and 2 for $n_2$ in equation (5).

  $\begin{array}{c}\Delta E\left(n_1\to n_2\right)=-2.18\times {10}^{-18}\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(6\right)}^2}\right)\text{ J}\\ =-4.84\times {10}^{-19}\text{ J}\end{array}$

Substitute $4.84\times {10}^{-19}\text{ J}$ 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 (7).

  $\begin{array}{c}\text{λ}=\frac{\left(\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{ m/s}\right)}{4.84\times {10}^{-19}\text{ J}}\\ =\left(4.11\times {10}^{-7}\text{ m}\right)\left(\frac{\text{1 nm}}{{\text{10}}^{-9}\text{ m}}\right)\\ =411\text{ nm}\end{array}$