Chapter 27, Problem 43P

(a)

To determine

The longest wavelength associated with the Paschen series.

Answer to Problem 43P

The longest wavelength associated with the Paschen series is $\underset{\_}{1874\text{nm}}$.

Explanation of Solution

The transition with the least energy will produce a photon with longest wavelength. The transition with least energy is $n=4$ to $n=3$ transition.

Write the expression to find the energy of the transition from $n=4$ to $n=3$.

    $E=E_4-E_3$

Here,$E_4$ is the energy of the $n=4$ state and $E_3$ is the energy of the $n=3$ state.

Rewrite the above equation in terms of the initial energy.

    $E=\frac{E_1}{4^2}-\frac{E_1}{3^2}$                                                                                                             (I)

Here, $E_1$ is the ground state energy.

Write the expression to find the wavelength in terms of energy.

    $\lambda =\frac{hc}{E}$                                                                                                                     (II)

Here, $h$ is the Planck’s constant, $c$ is the speed of light and $\lambda$ is the wavelength.

Conclusion:

Substitute $-13.61\text{eV}$ for $E_1$ in equation (I).

    $\begin{array}{c}E=\frac{-13.61\text{eV}}{4^2}-\frac{-13.61\text{eV}}{3^2}\\ =0.6616\text{eV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.998\times {10}^8\text{m}/\text{s}$ for $c$ and $0.6616\text{eV}\left(\frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)$ for $E$ in the above equation.

    $\begin{array}{c}\lambda =\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m}/\text{s}\right)}{0.6616\text{eV}\left(\frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =1874\text{nm}\end{array}$

Therefore, the longest wavelength associated with the Paschen series is $\underset{\_}{1874\text{nm}}$.

(b)

To determine

The wavelength of the series limit of the Paschen series.

Answer to Problem 43P

The wavelength of the series limit of the Paschen series is $\underset{\_}{820.0\text{nm}}$.

Explanation of Solution

The transition with the series limit is the transition between $n=\infty$ to $n=3$ transition.

Write the expression to find the energy of the transition from $n=\infty$ to $n=3$.

    $E=E_{\infty }-E_3$

Here,$E_{\infty }$ is the energy of the $n=\infty$ state and $E_3$ is the energy of the $n=3$ state.

Rewrite the above equation in terms of the initial energy.

    $E=\frac{E_1}{\infty }-\frac{E_1}{3^2}$                                                                                                            (III)

Here, $E_1$ is the ground state energy.

Write the expression to find the wavelength in terms of energy.

    $\lambda =\frac{hc}{E}$                                                                                                                   (IV)

Here, $h$ is the Planck’s constant, $c$ is the speed of light and $\lambda$ is the wavelength.

Conclusion:

Substitute $-13.61\text{eV}$ for $E_1$ in equation (III).

    $\begin{array}{c}E=\frac{-13.61\text{eV}}{\infty }-\frac{-13.61\text{eV}}{3^2}\\ =0-\frac{-13.61\text{eV}}{3^2}\\ =1.5122\text{eV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.998\times {10}^8\text{m}/\text{s}$ for $c$ and $1.5122\text{eV}\left(\frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)$ for $E$ in the equation (IV).

    $\begin{array}{c}\lambda =\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m}/\text{s}\right)}{1.5122\text{eV}\left(\frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =820.0\text{nm}\end{array}$

Therefore, the wavelength of the series limit of the Paschen series is $\underset{\_}{820.0\text{nm}}$.

(c)

To determine

The part of the electromagnetic spectrum in which the Paschen series is present.

Answer to Problem 43P

The part of the electromagnetic spectrum in which the Paschen series is present is the IR.

Explanation of Solution

The range of the wavelengths of the Paschen series extends from $820.0\text{nm}$ to $1874\text{nm}$. This set of wavelengths corresponds to the infrared region of the electromagnetic spectrum.

Therefore, the part of the electromagnetic spectrum in which the Paschen series is present is the IR.