Chapter 4, Problem 4.42P

Interpretation Introduction

Interpretation:

Wavelength of the first five lines in Paschen series has to be calculated.

Explanation of Solution

Wavelength of First line:

Energy of photon emitted on relaxation from $n=4$ state to $n=3$ state in first line of Paschen series and is calculated as shown below.

    $\begin{array}{c}{E}_{photon}=\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{3^2}-\frac{1}{4^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{9}-\frac{1}{16}\right)\\ =0.1059\times {10}^{-18}\text{J}\end{array}$

Wavelength of photon is calculated as shown below.

    $\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{0.1059\times {10}^{-18}\text{J}}\\ =187.5739887\times {10}^{-8}\text{m}\\ =1875\times {10}^{-9}\text{m}\\ \text{=1875}\text{nm}\end{array}$

Therefore, the wavelength of photon is calculated as $\text{1875}\text{nm}$.

Wavelength of Second line:

Energy of photon emitted on relaxation from $n=5$ state to $n=3$ state in second line of Paschen series and is calculated as shown below.

    $\begin{array}{c}{E}_{photon}=\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{3^2}-\frac{1}{5^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{9}-\frac{1}{25}\right)\\ =0.1550\times {10}^{-18}\text{J}\end{array}$

Wavelength of photon is calculated as shown below.

    $\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{0.1550\times {10}^{-18}\text{J}}\\ =128.1553897\times {10}^{-8}\text{m}\\ =1281\times {10}^{-9}\text{m}\\ \text{=1281}\text{nm}\end{array}$

Therefore, the wavelength of photon is calculated as $\text{1281}\text{nm}$.

Wavelength of Third line:

Energy of photon emitted on relaxation from $n=6$ state to $n=3$ state in third line of Paschen series and is calculated as shown below.

    $\begin{array}{c}{E}_{photon}=\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{3^2}-\frac{1}{6^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{9}-\frac{1}{36}\right)\\ =0.1816\times {10}^{-18}\text{J}\end{array}$

Wavelength of photon is calculated as shown below.

    $\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{0.1816\times {10}^{-18}\text{J}}\\ =109.3837302\times {10}^{-8}\text{m}\\ =1094\times {10}^{-9}\text{m}\\ \text{=1094}\text{nm}\end{array}$

Therefore, the wavelength of photon is calculated as $\text{1094}\text{nm}$.

Wavelength of Fourth line:

Energy of photon emitted on relaxation from $n=7$ state to $n=3$ state in third line of Paschen series and is calculated as shown below.

    $\begin{array}{c}{E}_{photon}=\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{9}-\frac{1}{49}\right)\\ =0.1977\times {10}^{-18}\text{J}\end{array}$

Wavelength of photon is calculated as shown below.

    $\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{0.1977\times {10}^{-18}\text{J}}\\ =100.4758998\times {10}^{-8}\text{m}\\ =1004\times {10}^{-9}\text{m}\\ \text{=1004}\text{nm}\end{array}$

Therefore, the wavelength of photon is calculated as $\text{1004}\text{nm}$.

Wavelength of Fifth line:

Energy of photon emitted on relaxation from $n=8$ state to $n=3$ state in third line of Paschen series and is calculated as shown below.

    $\begin{array}{c}{E}_{photon}=\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{3^2}-\frac{1}{8^2}\right)\\ =\left(2.1799\times {10}^{-18}\text{J}\right)\left(\frac{1}{9}-\frac{1}{64}\right)\\ =0.2082\times {10}^{-18}\text{J}\end{array}$

Wavelength of photon is calculated as shown below.

    $\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{0.2082\times {10}^{-18}\text{J}}\\ =95.40867147\times {10}^{-8}\text{m}\\ =954\times {10}^{-9}\text{m}\\ \text{=954}\text{nm}\end{array}$

Therefore, the wavelength of photon is calculated as $95.4\times {10}^{16}\text{m}$.

Summary of the wavelength of first five lines in Paschen series is given below.

$n_i$	4	5	6	7	8
$\text{Wavelength/nm}$	$1875$	$1281$	$1094$	$1004$	$954$