Chapter 11, Problem 11.67E

To four significant figures, the first four lines in the Balmer series in the hydrogen atom $\left(n_2=2\right)$ spectrum appear at $656.5$, $486.3$, $434.2$, and $410.3\text{nm}$. (a) From these numbers, calculate an average value of $R_{\text{H}}$, Rydberg constant. (b) At what wavelengths would similar transitions appear for ${\text{He}}^{+}$?

Interpretation Introduction

(a)

Interpretation:

The average value of $R_{\text{H}}$, the Rydberg constant is to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength or the energy difference of the lines present in the atomic spectrum of an element. The Rydberg equation for the hydrogen atom is represented as,

$\Delta E=R_{\text{H}}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Where,

• $R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $109737.31{\text{cm}}^{-1}$.

• $n_1$ and $n_2$ represents the energy levels.

• $\Delta E$ is the energy difference.

Answer to Problem 11.67E

The average value of $R_{\text{H}}$, the Rydberg constant is $2.18\times {10}^{-18}\text{J}$.

Explanation of Solution

For Balmer series in the hydrogen atom $n_2=2$. For the first four lines, that is, for $n_1=3,4,5$ and $6$ the spectrum appears at $656.5$, $486.3$, $434.2$, and $410.3\text{nm}$ respectively. The expression of wavelength using Rydberg constant is,

$\Delta E=\frac{hc}{\lambda }=R_{\text{H}}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Where,

• $R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $109737.31{\text{cm}}^{-1}$.

• $n_1$ and $n_2$ represents the energy levels.

• $\lambda$ is the wavelength for the transition from one from one energy level to another.

• $h$ is the Planck’s constant.

• $\Delta E$ is the energy difference.

Substitute the values of wavelength for $n_1=3$, the Rydberg constant in the given formula.

$\begin{array}{rll}{R}_{\text{H}}& =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{\left(656.5\times {10}^{-9}\text{m}\right)\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}\\ R_{\text{H}}& =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{\left(656.5\times {10}^{-9}\text{m}\right)\left(\frac{1}{4}-\frac{1}{9}\right)}\\ R_{\text{H}}& =& 2.18\times {10}^{-18}\text{J}\end{array}$

Substitute the values of wavelength for $n_1=4$, the Rydberg constant in the given formula.

$\begin{array}{rll}{R}_{\text{H}}& =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{\left(486.3\times {10}^{-9}\text{m}\right)\left(\frac{1}{2^2}-\frac{1}{4^2}\right)}\\ R_{\text{H}}& =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{\left(486.3\times {10}^{-9}\text{m}\right)\left(\frac{1}{4}-\frac{1}{16}\right)}\\ R_{\text{H}}& =& 2.18\times {10}^{-18}\text{J}\end{array}$

Substitute the values of wavelength for $n_1=5$, the Rydberg constant in the given formula.

$\begin{array}{rll}{R}_{\text{H}}& =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{\left(434.2\times {10}^{-9}\text{m}\right)\left(\frac{1}{2^2}-\frac{1}{5^2}\right)}\\ R_{\text{H}}& =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{\left(434.2\times {10}^{-9}\text{m}\right)\left(\frac{1}{4}-\frac{1}{25}\right)}\\ R_{\text{H}}& =& 2.18\times {10}^{-18}\text{J}\end{array}$

Substitute the values of wavelength for $n_1=6$, the Rydberg constant in the given formula.

$\begin{array}{rll}{R}_{\text{H}}& =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{\left(410.3\times {10}^{-9}\text{m}\right)\left(\frac{1}{2^2}-\frac{1}{6^2}\right)}\\ R_{\text{H}}& =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{\left(410.3\times {10}^{-9}\text{m}\right)\left(\frac{1}{4}-\frac{1}{36}\right)}\\ R_{\text{H}}& =& 2.18\times {10}^{-18}\text{J}\end{array}$

Since all the values of $R_{\text{H}}$ are the same, the average value of $R_{\text{H}}$, Rydberg constant is $2.18\times {10}^{-18}\text{J}$.

Conclusion

The average value of $R_{\text{H}}$, the Rydberg constant is $2.18\times {10}^{-18}\text{J}$.

Interpretation Introduction

(b)

Interpretation:

The wavelengths for which similar transitions occur for ${\text{He}}^{+}$ are to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength or the energy difference of the lines present in the atomic spectrum of an element. The Rydberg equation for the hydrogen atom is represented as,

$\Delta E=R_{\text{H}}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Where,

• $R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $109737.31{\text{cm}}^{-1}$.

• $n_1$ and $n_2$ represents the energy levels.

• $\Delta E$ is the energy difference.

Answer to Problem 11.67E

The wavelength of transition for helium cation for $n_1=3,4,5$ and $6$ the spectrum appears at $164.125\text{nm}$, $121.575\text{nm}$, $108.550\text{nm}$, and $102.575\text{nm}$ respectively.

Explanation of Solution

For Balmer series in the hydrogen atom $n_2=2$. For the first four lines, that is, for $n_1=3,4,5$ and $6$ the spectrum appears at $656.5$, $486.3$, $434.2$, and $410.3\text{nm}$.

For hydrogen atom the energy difference is calculated by,$\Delta E_{\text{H}}=\frac{e^4\mu }{8{\in }_0^2{h}^2}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Where,

• $\mu$ is the reduced mass.

• $e$ is the charge on electron.

• ${\in }_0$permittivity of free space.

• $h$ is the Planck’s constant.

• $n_1$ and $n_2$ represents the energy levels.

The energy difference using Rydberg constant for atoms other than hydrogen is calculated by,$\Delta E=\frac{Z^2{e}^4\mu }{8{\in }_0^2{h}^2}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Where,

• $Z$ is the atomic number.

• $\mu$ is the reduced mass.

• $e$ is the charge on electron.

• ${\in }_0$ is the permittivity of free space.

• $h$ is the Planck’s constant.

• $n_1$ and $n_2$ represents the energy levels.

For helium atom, $Z=2$.

Thus, relation between the energy difference of hydrogen and helium cation is,

$\begin{array}{l}\Delta E_{\text{He}}=2^2\Delta E_{\text{H}}\\ \Delta E_{\text{He}}=4\Delta E_{\text{H}}\end{array}$

The relation between the wavelengths of transition of hydrogen and helium cation is,

$\frac{hc}{{\lambda }_{\text{He}}}=4\frac{hc}{{\lambda }_{\text{H}}}$

${\lambda }_{\text{He}}=\frac{{\lambda }_{\text{H}}}{4}$ …(1)

Substitute the value of wavelength for $n_1=3$, in equation (1).

$\begin{array}{l}{\lambda }_{\text{He}}=\frac{656.5\text{nm}}{4}\\ {\lambda }_{\text{He}}=164.125\text{nm}\end{array}$

Substitute the value of wavelength for $n_1=4$, in equation (1).

$\begin{array}{l}{\lambda }_{\text{He}}=\frac{486.3\text{nm}}{4}\\ {\lambda }_{\text{He}}=121.575\text{nm}\end{array}$

Substitute the value of wavelength for $n_1=5$, in equation (1).

$\begin{array}{l}{\lambda }_{\text{He}}=\frac{434.2\text{nm}}{4}\\ {\lambda }_{\text{He}}=108.55\text{nm}\end{array}$

Substitute the value of wavelength for $n_1=5$, in equation (1).

$\begin{array}{l}{\lambda }_{\text{He}}=\frac{410.3\text{nm}}{4}\\ {\lambda }_{\text{He}}=102.575\text{nm}\end{array}$

Thus, the wavelength of transition for helium cation for $n_1=3,4,5$ and $6$ the spectrum appears at $164.125\text{nm}$, $121.575\text{nm}$, $108.550\text{nm}$, and $102.575\text{nm}$ respectively.

Conclusion

The wavelength of transition for helium cation for $n_1=3,4,5$ and $6$ the spectrum appears at $164.125\text{nm}$, $121.575\text{nm}$, $108.550\text{nm}$, and $102.575\text{nm}$ respectively.