# What will be the ratio of the wavelength of the first line to that of the second line of paschen series of H atom? A)256:175 B)175:256 C)15:16 D)24:27

Sep 30, 2017

The answer is (A) $256 : 175$

#### Explanation:

Your tool of choice here will be the Rydberg equation, which tells you the wavelength, $l a m \mathrm{da}$, of the photon emitted by an electron that makes a ${n}_{i} \to {n}_{f}$ transition in a hydrogen atom.

$\frac{1}{l a m \mathrm{da}} = R \cdot \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

Here

• $R$ is the Rydberg constant, equal to $1.097 \cdot {10}^{7}$ ${\text{m}}^{- 1}$
• ${n}_{i}$ is the initial energy level of the electron
• ${n}_{f}$ is the final energy level of the electron

Now, the Paschen series is characterized by ${n}_{f} = 3$. The first transition in the Paschen series corresponds to

${n}_{i} = 4 \text{ " -> " } {n}_{f} = 3$

In this transition, the electron drops from the fourth energy level to the third energy level.

You will have

$\frac{1}{l a m {\mathrm{da}}_{1}} = R \cdot \left(\frac{1}{3} ^ 2 - \frac{1}{4} ^ 2\right)$

The second transition in the Paschen series corresponds to

${n}_{i} = 5 \text{ " -> " } {n}_{f} = 3$

This time, you have

$\frac{1}{l a m {\mathrm{da}}_{2}} = R \cdot \left(\frac{1}{3} ^ 2 - \frac{1}{5} ^ 2\right)$

Now, to get the ratio of the first line to that of the second line, you need to divide the second equation by the first one.

$\frac{\frac{1}{l a m {\mathrm{da}}_{2}}}{\frac{1}{l a m {\mathrm{da}}_{1}}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{R}}} \cdot \left(\frac{1}{3} ^ 2 - \frac{1}{4} ^ 2\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{R}}} \cdot \left(\frac{1}{3} ^ 2 - \frac{1}{5} ^ 2\right)}$

This will be equivalent to

$\frac{l a m {\mathrm{da}}_{1}}{l a m {\mathrm{da}}_{2}} = \frac{\frac{1}{9} - \frac{1}{25}}{\frac{1}{9} - \frac{1}{16}}$

$\frac{l a m {\mathrm{da}}_{1}}{l a m {\mathrm{da}}_{2}} = \frac{\frac{25 - 9}{25 \cdot 9}}{\frac{16 - 9}{16 \cdot 9}} = \frac{16}{25 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}} \cdot \frac{16 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}}{7} = {16}^{2} / \left(25 \cdot 7\right)$

Therefore, you can say that you have

$\frac{l a m {\mathrm{da}}_{1}}{l a m {\mathrm{da}}_{2}} = \frac{256}{175}$