Chapter 7, Problem 7.80P

(a)

Interpretation Introduction

Interpretation:

Whether the $n=1$ range of wavelengths series for the hydrogen atom overlaps with the range $n=2$ wavelength series 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 7.80P

The $n_1=1$ range of wavelengths series for the hydrogen atom does not overlap with the range $n_1=2$ wavelength series

Explanation of Solution

The overlap between $n_1=1$ and $n_1=2$ series is checked by comparing the longest wavelength for $n_1=1$ and the shortest wavelength for $n_1=2$.

The longest wavelength in the $n_1=1$ series has $n_2=2$.

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$,1 for $n_1$ and 2 for $n_2$ in equation (1).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{1}-\frac{1}{4}\right)\\ =0.822582\times {10}^7{\text{ m}}^{-1}\\ \lambda =1.215\times {10}^{-7}\text{ m}\end{array}$

The shortest wavelength in the $n_1=2$ series has $n_2=\infty$.

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$,2 for $n_1$ and $\infty$ for $n_2$ in equation (1).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{2^2}-\frac{1}{\infty }\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{4}-0\right)\\ =0.274194\times {10}^7{\text{ m}}^{-1}\\ \lambda =3.647\times {10}^{-7}\text{ m}\end{array}$

The longest wavelength for $n_1=1$ series is shorter than the shortest wavelength for $n_1=2$ series. Hence, the overlap is not possible.

Conclusion

The $n=1$ range of wavelengths series for the hydrogen atom does not overlap with the range $n=2$ wavelength series

(b)

Interpretation Introduction

Interpretation:

Whether the $n=3$ range of wavelengths series for the hydrogen atom overlaps with the range $n=4$ wavelength series 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.

Answer to Problem 7.80P

The $n=4$ range of wavelengths series for the hydrogen atom does overlap with the range $n=3$ wavelength series

Explanation of Solution

The overlap between $n_1=3$ and $n_1=4$ series is checked by comparing the longest wavelength for $n_1=3$ and the shortest wavelength for $n_1=4$.

The longest wavelength in the $n_1=3$ series has $n_2=4$.

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$, 3 for $n_1$ and 4 for $n_2$ in equation (1).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{4^2}\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{9}-\frac{1}{16}\right)\\ =0.0533155\times {10}^7{\text{ m}}^{-1}\\ \lambda =1.875627\times {10}^{-6}\text{ m}\end{array}$

The shortest wavelength in the $n_1=4$ series has $n_2=\infty$.

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$,4 for $n_1$ and $\infty$ for $n_2$ in equation (1).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{4^2}-\frac{1}{\infty }\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{16}-0\right)\\ =0.0685485\times {10}^7{\text{ m}}^{-1}\\ \lambda =1.458821\times {10}^{-7}\text{ m}\end{array}$

The longest wavelength for $n_1=3$ series is longer than the shortest wavelength for $n_1=4$ series. Hence, the overlap is possible.

Conclusion

The $n=4$ range of wavelengths series for the hydrogen atom does overlap with the range $n=3$ wavelength series

(c)

Interpretation Introduction

Interpretation:

The number of lines in the $n_1=4$ series that lie in the range of the $n_1=5$ series 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.

Answer to Problem 7.80P

The number of lines in the $n_1=4$ series that lie in the range of the $n_1=5$ series is 2.

Explanation of Solution

The shortest wavelength in the $n_1=5$ series has $n_2=\infty$.

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$,5 for $n_1$ and $\infty$ for $n_2$ in equation (1).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{5^2}-\frac{1}{\infty }\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{25}-0\right)\\ =0.04387104\times {10}^7{\text{ m}}^{-1}\\ \lambda =2.2794\times {10}^{-7}\text{ m}\end{array}$

In the $n_1=4$ series, a few longest wavelengths are calculated to determine whether overlap is possible with the shortest wavelength for $n_1=5$.

The longest wavelength in the $n_1=4$ series has $n_2=5$.

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$, 4 for $n_1$ and 5 for $n_2$ in equation (1).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{4^2}-\frac{1}{5^2}\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{16}-\frac{1}{25}\right)\\ =0.02467746\times {10}^7{\text{ m}}^{-1}\\ \lambda =4.0522\times {10}^{-6}\text{ m}\end{array}$

For $n_1=4$ and $n_2=6$,

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$, 4 for $n_1$ and 6 for $n_2$ in equation (1).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{4^2}-\frac{1}{6^2}\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{16}-\frac{1}{36}\right)\\ =0.0381678\times {10}^7{\text{ m}}^{-1}\\ \lambda =2.6200095\times {10}^{-6}\text{ m}\end{array}$

For $n_1=4$ and $n_2=7$,

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$, 4 for $n_1$ and 7 for $n_2$ in equation (1).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{16}-\frac{1}{49}\right)\\ =0.0461742\times {10}^7{\text{ m}}^{-1}\\ \lambda =2.165711\times {10}^{-6}\text{ m}\end{array}$

The first two lines with $n_2$ values of 5 and 6 overlap with the shortest wavelength in the $n_1=5$ series.

Conclusion

The number of lines in the $n_1=4$ series that lie in the range of the $n_1=5$ series is 2.

(d)

Interpretation Introduction

Interpretation:

The implication about the hydrogen atom line spectrum made by the overlap at longer wavelengths 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.

Answer to Problem 7.80P

The implication about the hydrogen atom line spectrum made by the overlap at longer wavelengths is that the hydrogen spectrum becomes more complex at longer wavelengths.

Explanation of Solution

The longer wavelengths of a series have more overlapping with the short wavelengths of the successive series. The overlapping of the lines leads to the formation of a continuous spectrum in the form of band. This makes it difficult for the analyst to interpret the needed information. Hence, the overlapping of the lines leads to the complexity of the hydrogen spectrum.

Conclusion

The implication about the hydrogen atom line spectrum made by the overlap at longer wavelengths is that the hydrogen spectrum becomes more complex at longer wavelengths.