Chapter 3, Problem 12P

To determine

The four largest wavelengths for the Brackett and Pfund series of hydrogen.

Answer to Problem 12P

The four wavelengths for the Bracket series are $4.\text{055}\text{μm},2.627\text{μm},2.167\text{μm}\text{and}1.946\text{μm}$ and for the Pfund series is $7.46\text{μm},4.657\text{μm},3.743\text{μm}\text{and}3.299\text{μm}$.

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 Brackett series of hydrogen spectrum is $4$. So, the value for four brightest emission will be $5,6,7\&8$. In the same manner, the Pfund series lies at number $5$. So, the value of $k$ for the next four wavelength will be $6,7,8\&9$.

Conclusion:

For Brackett series:

Substitute $4$ for $n$, $5$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{4^2}-\frac{1}{5^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(44.44\right)\text{m}\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =4.\text{055}\text{μm}\end{array}$

Substitute $4$ for $n$, $6$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{4^2}-\frac{1}{6^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(28.8\right)\text{m}\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =2.627\text{μm}\end{array}$

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

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(23.75\right)\text{m}\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =2.167\text{μm}\end{array}$

Substitute $4$ for $n$, $8$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{4^2}-\frac{1}{8^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(21.33\right)\text{m}\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =1.946\text{μm}\end{array}$

For Pfund series:

Substitute $5$ for $n$, $6$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{5^2}-\frac{1}{6^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(81.82\right)\text{m}\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =7.46\text{μm}\end{array}$

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

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{5^2}-\frac{1}{7^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(51.04\right)\text{m}\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =4.657\text{μm}\end{array}$

Substitute $5$ for $n$, $8$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{5^2}-\frac{1}{8^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(41.02\right)\text{m}\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =3.743\text{μm}\end{array}$

Substitute $5$ for $n$, $9$ for $k$ and $1.096\times {10}^7{\text{m}}^{-1}$ for $R_H$ in equation (I).

  $\begin{array}{c}\frac{1}{\lambda }=\left(1.096\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1}{5^2}-\frac{1}{9^2}\right)\\ \lambda =\frac{1}{1.096\times {10}^7{\text{m}}^{-1}}\left(36.16\right)\text{m}\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =3.299\text{μm}\end{array}$

Thus, the four wavelengths for the Bracket series are $4.\text{055}\text{μm},2.627\text{μm},2.167\text{μm}\text{and}1.946\text{μm}$ and for the Pfund series is $7.46\text{μm},4.657\text{μm},3.743\text{μm}\text{and}3.299\text{μm}$.