Chapter 3, Problem 15P

(a)

To determine

The wavelengths that lies in the range of $980\text{nm}-192\text{0}\text{nm}$ and the name of the series to which they belong.

Answer to Problem 15P

The wavelength that lie in the range of $980\text{nm}-192\text{0}\text{nm}$ lies in the Balmer series and the wavelengths are $955.4\text{nm},1005.9\text{nm},1094.8\text{nm},1283.0\text{nm}\&1876.8\text{nm}$.

Explanation of Solution

Write the expression for the Rydberg’s equation.

    $\frac{1}{\lambda }=R_H\left(\frac{1}{n^2}-\frac{1}{k^2}\right)$        (I)

Here, $\lambda$ is the smallest wavelength of emission, $R_H$ is the Rydberg’s constant, $n$ is the number of the series of emission and $k$ is the next level up to which the emission occurs.

The number corresponding to the Paschen series of hydrogen spectrum is $3$. So, the value for four brightest emission will be $4,5,6,7\&8$.

Conclusion:

For Paschen series:

Substitute $3$ for $n$, $4$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{4^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(20.57\right)\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =1876.8\text{nm}\end{array}$

Substitute $3$ for $n$, $5$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{5^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(14.06\right)\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =1283.0\text{nm}\end{array}$

Substitute $3$ for $n$, $6$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{6^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(12\right)\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =1094.8\text{nm}\end{array}$

Substitute $3$ for $n$, $7$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(11.025\right)\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =1005.9\text{nm}\end{array}$

Substitute $3$ for $n$, $8$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{8^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(10.472\right)\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =955.4\text{nm}\end{array}$

Thus, the wavelength that lie in the range of $980\text{nm}-192\text{0}\text{nm}$ lies in the Balmer series and the wavelengths are $955.4\text{nm},1005.9\text{nm},1094.8\text{nm},1283.0\text{nm}\&1876.8\text{nm}$.

(b)

To determine

The reason for the visibility of the only three wavelengths by the detector.

Answer to Problem 15P

The three wavelengths are detected by the detector because the detector is moving away from the source and other wavelength is not visible for the detector.

Explanation of Solution

The wavelengths lying in the visible region are the four wavelengths that should be detected by the detector used. Since the detector is moving away from the source, there is a possibility that the higher wavelengths might have undergone red shift.

Due to red shift, the wavelength would have increased and might have gone out of the visible range of the electromagnetic spectrum.

Conclusion:

Thus, the three wavelengths are detected by the detector because the detector is moving away from the source and other wavelength is not visible for the detector.

(c)

To determine

The speed of the stellar object that emits the spectrum.

Answer to Problem 15P

The speed of the moving stellar object is $1.2\times {10}^7\text{m}/\text{s}$.

Explanation of Solution

Write the expression for the relation between the observed wavelength and the wavelength from the source.

  $\frac{{\lambda }_{observed}}{{\lambda }_{source}}=\sqrt{\frac{1+\beta }{1-\beta }}$

Simplify and rearrange the above expression for $\beta$.

  $\beta =\frac{{\left(\frac{{\lambda }_{observed}}{{\lambda }_{source}}\right)}^2-1}{{\left(\frac{{\lambda }_{observed}}{{\lambda }_{source}}\right)}^2+1}$        (II)

Here, ${\lambda }_{observed}$ is the observed wavelength, ${\lambda }_{source}$ is the wavelength emitted by the source and $\beta$ is the relativistic factor.

Write the expression for the speed of the stellar object.

  $v=\beta \cdot c$        (III)

Here, $v$ is the speed of the object and $c$ is the speed of light.

Conclusion:

Substitute $1334.5\text{nm}$ for ${\lambda }_{oberved}$ and $1283.0\text{nm}$ for ${\lambda }_{source}$ in equation (II).

  $\begin{array}{c}\beta =\frac{{\left(\frac{1334.5\text{nm}}{1283.0\text{nm}}\right)}^2-1}{{\left(\frac{1334.5\text{nm}}{1283.0\text{nm}}\right)}^2+1}\\ =\frac{0.0818}{2.0818}\\ =0.0392\end{array}$

Substitute $0.0392$ for $\beta$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (III).

  $\begin{array}{c}v=\left(0.0392\right)\left(3\times {10}^8\text{m}/\text{s}\right)\\ \approx 1.2\times {10}^7\text{m}/\text{s}\end{array}$

Thus, the speed of the moving stellar object is $1.2\times {10}^7\text{m}/\text{s}$.