Chapter 7, Problem 7.68P

Interpretation Introduction

Interpretation:

The $n_{\text{initial}}$ values for each of the five lines in the hydrogen atom spectrum are 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 the wavelength from $\stackrel{\text{o}}{\text{A}}$ to $\text{m}$ is as follows:

$1\stackrel{\text{o}}{\text{A}}=1\times {10}^{-10}\text{ m}$

Answer to Problem 7.68P

For the lines of wavelengths $1212.7\stackrel{\text{o}}{\text{A}}$, $4340.5\stackrel{\text{o}}{\text{A}}$, $4861.3\stackrel{\text{o}}{\text{A}}$, $6562.8\stackrel{\text{o}}{\text{A}}$ and $10938\stackrel{\text{o}}{\text{A}}$, the values of $n_{\text{initial}}$ are 2, 5, 4, 3 and 6 respectively.

Explanation of Solution

The radiation of wavelength $1212.7\stackrel{\text{o}}{\text{A}}$ falls within the ultraviolet region of the electromagnetic spectrum. The ultraviolet series of emission lines are obtained when the value of $n_{\text{final}}$ in the Rydberg’s constant is 1.

Substitute $1212.7\stackrel{\text{o}}{\text{A}}$ for $\lambda$, $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$ and 1 for $n_1$ in equation (1).

$\begin{array}{c}\frac{1}{1212.7\stackrel{\text{o}}{\text{A}}}\left(\frac{{10}^{-10}\text{ m}}{1\stackrel{\text{o}}{\text{A}}}\right)=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right)\\ \frac{1}{1.2127\times {10}^{-7}\text{ m}}=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right)\\ 0.7518456=\left(1-\frac{1}{n_2^2}\right)\end{array}$

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

$\begin{array}{c}\left(\frac{1}{n_2^2}\right)=1-0.7518456\\ \left(\frac{1}{n_2^2}\right)=0.2481544\\ n_2^2=4.029749\\ n_2=2\end{array}$

Thus, for the line of wavelength $1212.7\stackrel{\text{o}}{\text{A}}$, $n_{\text{initial}}$ is 2.

The radiation of wavelengths $4340.5\stackrel{\text{o}}{\text{A}}$, $4861.3\stackrel{\text{o}}{\text{A}}$ and $6562.8\stackrel{\text{o}}{\text{A}}$ fall within the visible region of the electromagnetic spectrum. The visible series of emission lines are obtained when the value of $n_{\text{final}}$ in the Rydberg’s constant is 2.

Substitute $4340.5\stackrel{\text{o}}{\text{A}}$ for $\lambda$, $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$ and 2 for $n_1$ in equation (1).

$\begin{array}{c}\frac{1}{4340.5\stackrel{\text{o}}{\text{A}}}\left(\frac{{10}^{-10}\text{ m}}{1\stackrel{\text{o}}{\text{A}}}\right)=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{2^2}-\frac{1}{n_2^2}\right)\\ \frac{1}{4.3405\times {10}^{-7}\text{ m}}=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{4}-\frac{1}{n_2^2}\right)\\ 0.2100594=\left(\frac{1}{4}-\frac{1}{n_2^2}\right)\end{array}$

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

$\begin{array}{c}\left(\frac{1}{n_2^2}\right)=\frac{1}{4}-0.2100594\\ \left(\frac{1}{n_2^2}\right)=0.0399406\\ n_2^2=25.037180\\ n_2=5\end{array}$

Thus, for the line of wavelength $4340.5\stackrel{\text{o}}{\text{A}}$, $n_{\text{initial}}$ is 5.

Substitute $4861.3\stackrel{\text{o}}{\text{A}}$ for $\lambda$, $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$ and 2 for $n_1$ in equation (1).

$\begin{array}{c}\frac{1}{4861.3\stackrel{\text{o}}{\text{A}}}\left(\frac{{10}^{-10}\text{ m}}{1\stackrel{\text{o}}{\text{A}}}\right)=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{2^2}-\frac{1}{n_2^2}\right)\\ \frac{1}{4.8613\times {10}^{-7}\text{ m}}=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{4}-\frac{1}{n_2^2}\right)\\ 0.1875554=\left(\frac{1}{4}-\frac{1}{n_2^2}\right)\end{array}$

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

$\begin{array}{c}\left(\frac{1}{n_2^2}\right)=\frac{1}{4}-0.1875554\\ \left(\frac{1}{n_2^2}\right)=0.0624446\\ n_2^2=16.0141949\\ n_2=4\end{array}$

Thus, for the line of wavelength $4861.3\stackrel{\text{o}}{\text{A}}$, $n_{\text{initial}}$ is 4.

Substitute $6562.8\stackrel{\text{o}}{\text{A}}$ for $\lambda$, $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$ and 2 for $n_1$ in equation (1).

$\begin{array}{c}\frac{1}{6562.8\stackrel{\text{o}}{\text{A}}}\left(\frac{{10}^{-10}\text{ m}}{1\stackrel{\text{o}}{\text{A}}}\right)=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{2^2}-\frac{1}{n_2^2}\right)\\ \frac{1}{6.5628\times {10}^{-7}\text{ m}}=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{4}-\frac{1}{n_2^2}\right)\\ 0.1523739=\left(\frac{1}{4}-\frac{1}{n_2^2}\right)\end{array}$

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

$\begin{array}{c}\left(\frac{1}{n_2^2}\right)=\frac{1}{4}-0.1523739\\ \left(\frac{1}{n_2^2}\right)=0.0976261\\ n_2^2=10.2430\\ n_2=3\end{array}$

Thus, for the line of wavelength $6562.8\stackrel{\text{o}}{\text{A}}$, $n_{\text{initial}}$ is 3.

The radiation of wavelength $10938\stackrel{\text{o}}{\text{A}}$ falls within the infrared region of the electromagnetic spectrum. The infrared series of emission lines are obtained when the value of $n_{final}$ in the Rydberg’s constant is 3.

Substitute $10938\stackrel{\text{o}}{\text{A}}$ for $\lambda$, $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$ and 3 for $n_1$ in equation (1).

$\begin{array}{c}\frac{1}{10938\stackrel{\text{o}}{\text{A}}}\left(\frac{{10}^{-10}\text{ m}}{1\stackrel{\text{o}}{\text{A}}}\right)=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{n_2^2}\right)\\ \frac{1}{10.938\times {10}^{-7}\text{ m}}=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{9}-\frac{1}{n_2^2}\right)\\ 0.0833573=\left(\frac{1}{9}-\frac{1}{n_2^2}\right)\end{array}$

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

$\begin{array}{c}\left(\frac{1}{n_2^2}\right)=\frac{1}{9}-0.0833573\\ \left(\frac{1}{n_2^2}\right)=0.0277538\\ n_2^2=36.031102\\ n_2=6\end{array}$

Thus, for the line of wavelength $10938\stackrel{\text{o}}{\text{A}}$, $n_{\text{initial}}$ is 6.

Conclusion

For the lines of wavelengths $1212.7\stackrel{\text{o}}{\text{A}}$, $4340.5\stackrel{\text{o}}{\text{A}}$, $4861.3\stackrel{\text{o}}{\text{A}}$, $6562.8\stackrel{\text{o}}{\text{A}}$ and $10938\stackrel{\text{o}}{\text{A}}$, the values of $n_{\text{initial}}$ are 2, 5, 4, 3 and 6 respectively.