Chapter 4, Problem 26P

To determine

The isotope shifts for each of the four visible Balmer series wavelengths for deuterium and tritium compared with hydrogen.

Answer to Problem 26P

The isotope shifts for each of the four visible Balmer series wavelengths for deuterium and tritium compared with hydrogen is calculated below.

Explanation of Solution

The isotope shift of the spectral lines is defined as the shift in wavelength sue to different isotopic mass of given elements.

Value of Rydberg constant for atomic hydrogen is $0.99946R_{\infty }$ , for deuterium is  $0.99973R_{\infty }$ for tritium is $0.99982R_{\infty }$.

Here, $R_{\infty }$ Rydberg constant for infinite nuclear mass and its value is $1.097373\times {10}^7{m}^{-1}$ .

Lower state of Balmer series is $n_l=2$, The symbol $\alpha$ refers to $n_u=3$, $\beta$ refers to $n_u=4$, $\gamma$ refers to $n_u=5$  and $\delta$ refers to $n_u=6$.

Write the expression for isotope shifted wavelengths.

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

Here, $\lambda$ is the wavelength, $n_l$ is the number of lower energy state and $n_u$ is the number of upper energy state.

Wavelength of $H_{\alpha }$ line is calculated when $n_l=2$ and $n_u=3$.

    $\begin{array}{c}\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\\ =0.13889R\end{array}$        (I)

Wavelength of $H_{\beta }$ line is calculated when $n_l=2$ and $n_u=4$.

    $\begin{array}{c}\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\\ =0.1875R\end{array}$        (II)

Wavelength of $H_{\gamma }$ line is calculated when $n_l=2$ and $n_u=5$.

    $\begin{array}{c}\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{5^2}\right)\\ =0.21R\end{array}$        (III)

Wavelength of $H_{\delta }$ line is calculated when $n_l=2$ and $n_u=6$.

    $\begin{array}{c}\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{6^2}\right)\\ =0.222R\end{array}$        (IV)

Conclusion:

Value of Rydberg constant for atomic hydrogen is $0.99946R_{\infty }$ , for deuterium is  $0.99973R_{\infty }$ for tritium is $0.99982R_{\infty }$.

 For $\lambda \left(H_{\alpha },\text{hydrogen}\right)$:

Substitute $0.99946R_{\infty }$ for $R$ and $1.097373\times {10}^7{m}^{-1}$   for $R_{\infty }$ in equation (I).   

    $\begin{array}{c}\frac{1}{\lambda }=0.13889\left(0.99946R_{\infty }\right)\\ =0.13889\left(0.99946\left(1.097373\times {10}^7{\text{m}}^{\text{-1}}\right)\right)\\ =1523318.3{\text{m}}^{\text{-1}}\\ \lambda =656.47\text{nm}\end{array}$

Similarly value of $\lambda \left(H_{\alpha },\text{deuterium}\right)=656.29\text{nm}$ and value of $\lambda \left(H_{\alpha },\text{tritium}\right)=656.23\text{nm}$

For $\lambda \left(H_{\beta },\text{hydrogen}\right)$:

Substitute $0.99946R_{\infty }$ for $R$ and $1.097373\times {10}^7{m}^{-1}$   for $R_{\infty }$ in equation (II).   

    $\begin{array}{c}\frac{1}{\lambda }=0.1875\left(0.99946R_{\infty }\right)\\ =0.1875\left(0.99946\left(1.097373\times {10}^7{\text{m}}^{\text{-1}}\right)\right)\\ =2056463.3{\text{m}}^{\text{-1}}\\ \lambda =486.27\text{nm}\end{array}$

Similarly value of $\lambda \left(H_{\beta },\text{deuterium}\right)=486.14\text{nm}$ and value of $\lambda \left(H_{\beta },\text{tritium}\right)=486.10\text{nm}$.

For $\lambda \left(H_{\gamma },\text{hydrogen}\right)$:

Substitute $0.99946R_{\infty }$ for $R$ and $1.097373\times {10}^7{m}^{-1}$   for $R_{\infty }$ in equation (II).   

    $\begin{array}{c}\frac{1}{\lambda }=0.21\left(0.99946R_{\infty }\right)\\ =0.21\left(0.99946\left(1.097373\times {10}^7{\text{m}}^{\text{-1}}\right)\right)\\ =2303238.88{\text{m}}^{\text{-1}}\\ \lambda =434.17\text{nm}\end{array}$

Similarly value of $\lambda \left(H_{\gamma },\text{deuterium}\right)=434.05\text{nm}$ and value of $\lambda \left(H_{\gamma },\text{tritium}\right)=434.02\text{nm}$.

For $\lambda \left(H_{\delta },\text{hydrogen}\right)$:

Substitute $0.99946R_{\infty }$ for $R$ and $1.097373\times {10}^7{m}^{-1}$   for $R_{\infty }$ in equation (II).   

    $\begin{array}{c}\frac{1}{\lambda }=0.22\left(0.99946R_{\infty }\right)\\ =0.22\left(0.99946\left(1.097373\times {10}^7{\text{m}}^{\text{-1}}\right)\right)\\ =2412916.9{\text{m}}^{\text{-1}}\\ \lambda =410.29\text{nm}\end{array}$

Similarly value of $\lambda \left(H_{\delta },\text{deuterium}\right)=410.18\text{nm}$ and value of $\lambda \left(H_{\delta },\text{tritium}\right)=410.15\text{nm}$

Thus, the isotope shifts for each of the four visible Balmer series wavelengths for deuterium and tritium compared with hydrogen is calculated above.