Chapter 15, Problem 15.4E

Determine the value of $R$ for deuterium $\left({}^{\text{2}}\text{H}\right)$ and tritium $\left({}^3\text{H}\right)$ using equation 15.3. You will have to look up the masses of the $\text{D}$ and $\text{T}$ nuclei. By what percentage do these $R$ values differ from that for hydrogen?

Interpretation Introduction

Interpretation:

The value of $R$ for deuterium and tritium is to be calculated. The percentage by which these values differ from that of hydrogen is to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength of the lines present in the atomic spectrum of an element. The Rydberg constant is calculated by the expression shown below.

$R=\frac{e^4\mu }{8{\in }_{\text{o}}^{2}c{h}^3}$

Answer to Problem 15.4E

The values of $R$ for deuterium and tritium are $109586.5{\text{cm}}^{-1}$ and $109596.14{\text{cm}}^{-1}$, respectively. The percentage difference between Rydberg constant of hydrogen and Rydberg constant of deuterium is $0.083\%$, and the percentage difference between Rydberg constant of hydrogen and Rydberg constant of tritium is $0.074\%$.

Explanation of Solution

The mass of deuterium nuclei is $3.347\times {10}^{-27}\text{kg}$. The mass of tritium nuclei is $5.022\times {10}^{-27}\text{kg}$.

The Rydberg constant is calculated by the expression given below.

$R=\frac{e^4\mu }{8{\in }_{\text{o}}^{2}c{h}^3}$…(1)

Where,

• $h$ is the Planck’s constant.

• $\mu$ is the reduced mass.

• ${\in }_{\text{o}}$ is the permittivity.

• $e$ is the charge of the electron.

• $c$ is the speed of light.

The reduced mass is calculated by the expression given below.

$\mu =\frac{m_e\times m_N}{m_e+m_N}$…(2)

Where,

• $m_e$ is the mass of electron.

• $m_N$ is the mass of the neutron.

• $\mu$ is the reduced mass.

The value of $m_e$ is $9.109\times {10}^{-31}\text{kg}$.

Substitute the values of $m_e$ and $m_N$ for the calculation of the reduced mass of deuterium in equation (2).

$\begin{array}{rll}\mu & =& \frac{9.109\times {10}^{-31}\text{kg}\times 3.347\times {10}^{-27}\text{kg}}{9.109\times {10}^{-31}\text{kg}+\left(3.347\times {10}^{-27}\text{kg}\right)}\\ & =& \frac{30.4878\times {10}^{-58}}{33479.109\times {10}^{-31}}\\ & =& 9.1065\times {10}^{-31}\text{kg}\end{array}$

The value of $e$ is $1.602\times {10}^{-19}\text{C}$.

The value of ${\in }_{\text{o}}$ is $8.854\times {10}^{-12}{\text{C}}^2{\text{s}}^{\text{2}}/\text{kg}{\text{m}}^2$.

The value of $h$ is $6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-1}$.

Substitute the values of $e$, ${\in }_{\text{o}}$, $h$ and $\mu$ in equation (1).

$\begin{array}{rll}R& =& \frac{{\left(1.602\times {10}^{-19}\text{C}\right)}^4\times \left(9.1065\times {10}^{-31}\text{kg}\right)}{8\times {\left(8.854\times {10}^{-12}{\text{C}}^2{\text{s}}^{\text{2}}/\text{kg}\cdot {\text{m}}^3\right)}^2{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}\cdot {\text{s}}^{-1}\right)}^3\left(3\times {10}^8\text{m}\cdot {\text{s}}^{-1}\right)}\\ & =& \frac{59.9793\times {10}^{-107}{\text{m}}^{-1}}{54.7324\times {10}^{-114}}\\ & =& 1.095865\times {10}^{+7}{\text{m}}^{-1}\end{array}$

The units of Rydberg constant can be represented in ${\text{cm}}^{-1}$ as shown below.

$\begin{array}{l}R=1.095865\times {10}^{+7}\times {10}^{-2}{\text{cm}}^{-1}\\ R=1.095865\times {10}^{+5}{\text{cm}}^{-1}\\ R=109586.5{\text{cm}}^{-1}\end{array}$

The value of Rydberg constant is $109586.5{\text{cm}}^{-1}$.

Substitute the values of $m_e$ and $m_N$ for the calculation of the reduced mass of tritium in equation (2).

$\begin{array}{rll}\mu & =& \frac{9.109\times {10}^{-31}\text{kg}\times 5.022\times {10}^{-27}\text{kg}}{9.109\times {10}^{-31}\text{kg}+\left(5.022\times {10}^{-27}\text{kg}\right)}\\ & =& \frac{45.7454\times {10}^{-58}}{50229.109\times {10}^{-31}}\\ & =& 9.1073\times {10}^{-31}\text{kg}\end{array}$

The value of $e$ is $1.602\times {10}^{-19}\text{C}$.

The value of ${\in }_{\text{o}}$ is $8.854\times {10}^{-12}{\text{C}}^2{\text{s}}^{\text{2}}/\text{kg}{\text{m}}^2$.

The value of $h$ is $6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-1}$.

Substitute the values of $e$, ${\in }_{\text{o}}$, $h$ and $\mu$ in equation (1).

$\begin{array}{rll}R& =& \frac{{\left(1.602\times {10}^{-19}\text{C}\right)}^4\times \left(9.1073\times {10}^{-31}\text{kg}\right)}{8\times {\left(8.854\times {10}^{-12}{\text{C}}^2{\text{s}}^{\text{2}}/\text{kg}\cdot {\text{m}}^3\right)}^2{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^{\text{2}}\cdot {\text{s}}^{-1}\right)}^3\left(3\times {10}^8\text{m}\cdot {\text{s}}^{-1}\right)}\\ & =& \frac{59.9846\times {10}^{-107}{\text{m}}^{-1}}{54.7324\times {10}^{-114}}\\ & =& 1.09596\times {10}^{+7}{\text{m}}^{-1}\end{array}$

The units of Rydberg constant can be represented in ${\text{cm}}^{-1}$ as shown below.

$\begin{array}{l}R=1.0959614\times {10}^{+7}\times {10}^{-2}{\text{cm}}^{-1}\\ R=1.0959614\times {10}^{+5}{\text{cm}}^{-1}\\ R=109596.14{\text{cm}}^{-1}\end{array}$

The value of Rydberg constant is $109596.14{\text{cm}}^{-1}$.

The Rydberg constant for hydrogen using the reduced mass is $109677.58{\text{cm}}^{-1}$.

The percentage difference is calculated by the ratio of the difference between the two Rydberg constants to Rydberg constant of hydrogen.

The calculation of percentage difference between Rydberg constant for deuterium and hydrogen is shown below.

$\%\text{Difference}=\frac{R_H-R_D}{R_H}\times 100$

Substitute the values of Rydberg constant of hydrogen and Rydberg constant of deuterium in the above formula.

$\begin{array}{rll}\%\text{Difference}& =& \frac{109677.58{\text{cm}}^{-1}-109586.5{\text{cm}}^{-1}}{109677.58{\text{cm}}^{-1}}\times 100\\ & =& 0.083\%\end{array}$

The percentage difference between the Rydberg constant of hydrogen and Rydberg constant of deuterium is $0.083\%$.

The calculation of percentage difference between the Rydberg constant for tritium and hydrogen is shown below.

$\%\text{Difference}=\frac{R_H-R_T}{R_H}\times 100$

Substitute the values of Rydberg constant of hydrogen and Rydberg constant of tritium in the above formula.

$\begin{array}{rll}\%\text{Difference}& =& \frac{109677.58{\text{cm}}^{-1}-109596.14{\text{cm}}^{-1}}{109677.58{\text{cm}}^{-1}}\times 100\\ & =& 0.074\%\end{array}$

The percentage difference between the Rydberg constant of hydrogen and Rydberg constant of tritium is $0.074\%$.

Conclusion

The values of $R$ for deuterium and tritium are $109586.5{\text{cm}}^{-1}$ and $109596.14{\text{cm}}^{-1}$, respectively. The percentage difference between the Rydberg constant of hydrogen and Rydberg constant of deuterium is $0.083\%$, and the percentage difference between the Rydberg constant of hydrogen and Rydberg constant of tritium is $0.074\%$.