Chapter 7, Problem 7.73P

(a)

Interpretation Introduction

Interpretation:

The energy level to which a ground state electron in a hydrogen atom jumps to after it absorbs a photon of wavelength $94.91\text{ nm}$ is to be determined.

Concept introduction:

An atom of hydrogen contains one electron. But the spectrum of hydrogen consists of a large number of lines. This is so because a sample of hydrogen contains a very large number of atoms. When energy is supplied to a sample of gaseous atoms of hydrogen, different atoms absorb different amounts of energy. Therefore, the electrons in different atoms jump to different energy levels. Upon losing the energies gained initially, the electrons jump back to lower energy levels and release radiations of different wavelengths.

The equation used to predict the position and wavelength of any line in a given series is called the Rydberg’s equation.

Rydberg’s equation is as follows:

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$        (1)

Here,

$\lambda$ is the wavelength of the line.

$n_1$ and $n_2$ are positive integers, with ${\text{n}}_2>n_1$.

$R$ is the Rydberg’s constant.

The conversion factor to convert wavelength from $\text{nm}$ to $\text{m}$ is,

$1\text{nm}=1\times {10}^{-9}\text{ m}$

Answer to Problem 7.73P

An electron in a hydrogen atom moves to energy level 5 from the ground state upon absorption of a photon of wavelength $94.91\text{ nm}$.

Explanation of Solution

In the ground state of a hydrogen atom, $n=1$.

The value of the Rydberg’s constant is $1.096776\times {10}^7{\text{ m}}^{-1}$.

Substitute 1 for $n_1$, $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$ and $94.91\text{ nm}$ for $\lambda$ in equation (1).

$\begin{array}{c}\frac{1}{\left(94.91\text{ nm}\right)\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)}=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right)\\ \frac{1}{9.491\times {10}^{-8}\text{ m}}=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(1-\frac{1}{n_2^2}\right)\\ 0.9606608=\left(1-\frac{1}{n_2^2}\right)\end{array}$

Rearrange the above equation and calculate the value of $n_2$ as follows:

$\begin{array}{c}\frac{1}{n_2^2}=1-0.9606608\\ \frac{1}{n_2^2}=0.0393392\\ n_2^2=24.41994\\ n_2=5\end{array}$

Conclusion

An electron in a hydrogen atom moves to energy level 5 from the ground state upon absorption of a photon of wavelength $94.91\text{ nm}$.

(b)

Interpretation Introduction

Interpretation:

The intermediate energy level to which the electron jumps after emission of a photon of wavelength $1281\text{ nm}$ is to be determined.

Concept introduction:

An atom of hydrogen contains one electron. But the spectrum of hydrogen consists of a large number of lines. This is so because a sample of hydrogen contains a very large number of atoms. When energy is supplied to a sample of gaseous atoms of hydrogen, different atoms absorb different amounts of energy. Therefore, the electrons in different atoms jump to different energy levels. Upon losing the energies gained initially, the electrons jump back to lower energy levels and release radiations of different wavelengths.

The equation used to predict the position and wavelength of any line in a given series is called the Rydberg’s equation.

Rydberg’s equation is as follows:

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$        (1)

Here,

$\lambda$ is the wavelength of the line.

$n_1$ and $n_2$ are positive integers, with ${\text{n}}_2>n_1$.

$R$ is the Rydberg’s constant.

The conversion factor to convert wavelength from $\text{nm}$ to $\text{m}$ is,

$1\text{nm}=1\times {10}^{-9}\text{ m}$

Answer to Problem 7.73P

The intermediate energy level to which the electron jumps after emission of a photon of wavelength $1281\text{ nm}$ is $n=3$.

Explanation of Solution

The emission of a photon leads to the transition of an electron from a higher energy level to a lower energy level. Initially, the electron was in energy level 5. Since in the Rydberg’s equation, ${\text{n}}_2>n_1$, therefore $n_2=5$.

The value of the Rydberg’s constant is $1.096776\times {10}^7{\text{ m}}^{-1}$.

Substitute 5 for ${\text{n}}_2$, $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$ and $1281\text{ nm}$ for $\lambda$ in equation (1).

$\begin{array}{c}\frac{1}{\left(1281\text{ nm}\right)\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)}=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{n_1^2}-\frac{1}{5^2}\right)\\ \frac{1}{1.281\times {10}^{-6}\text{ m}}=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{n_1^2}-\frac{1}{25}\right)\\ 0.07118=\left(\frac{1}{n_1^2}-\frac{1}{25}\right)\end{array}$

Rearrange the above equation and calculate the value of $n_1$ as follows:

$\begin{array}{c}\frac{1}{n_1^2}=0.07118+0.04000\\ \frac{1}{n_1^2}=0.11118\\ n_1^2=8.9944\\ n_1=3\end{array}$

Conclusion

The intermediate energy level to which the electron jumps after emission of a photon of wavelength $1281\text{ nm}$ is $n=3$.

(c)

Interpretation Introduction

Interpretation:

The wavelength of the photon emitted after the electron jumps from $n=3$ to $n=1$ is to be determined.

Concept introduction:

An atom of hydrogen contains one electron. But the spectrum of hydrogen consists of a large number of lines. This is so because a sample of hydrogen contains a very large number of atoms. When energy is supplied to a sample of gaseous atoms of hydrogen, different atoms absorb different amounts of energy. Therefore, the electrons in different atoms jump to different energy levels. Upon losing the energies gained initially, the electrons jump back to lower energy levels and release radiations of different wavelengths.

The equation used to predict the position and wavelength of any line in a given series is called the Rydberg’s equation.

Rydberg’s equation is as follows:

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$        (1)

Here,

$\lambda$ is the wavelength of the line.

$n_1$ and $n_2$ are positive integers, with ${\text{n}}_2>n_1$.

$R$ is the Rydberg’s constant.

The conversion factor to convert wavelength from $\text{nm}$ to $\text{m}$ is,

$1\text{nm}=1\times {10}^{-9}\text{ m}$

Answer to Problem 7.73P

The wavelength of the photon emitted after the electron jumps from $n=3$ to $n=1$ is $102.6\text{ nm}$.

Explanation of Solution

The emission of a photon leads to the transition of an electron from a higher energy level to a lower energy level. Initially, the electron was in energy level 3 and later jumps to energy level 1. Since in the Rydberg’s equation, ${\text{n}}_2>n_1$, therefore $n_2=3$.

The value of the Rydberg’s constant is $1.096776\times {10}^7{\text{ m}}^{-1}$.

Substitute 3 for ${\text{n}}_2$, 1 for $n_1$ and $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$ in equation (1).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\\ \frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(1-\frac{1}{9}\right)\\ \frac{1}{\lambda }=9.74912\times {10}^6{\text{ m}}^{-1}\\ \lambda =102.6\text{ nm}\end{array}$

Conclusion

The wavelength of the photon emitted after the electron jumps from $n=3$ to $n=1$ is $102.6\text{ nm}$.