Question

In Paschen series, wavelength of first line is '$\lambda_1$' and for Brackett series, wavelength of first line is '$\lambda_2$' then ratio $\frac{\lambda_1}{\lambda_2}$ is ______.

• A. $\frac{7}{400}$
• B. $\frac{9}{144}$
• C. $\frac{81}{175}$
• D. $\frac{108}{509}$

Hint:

Series bases: Lyman ($n=1$), Balmer ($n=2$), Paschen ($n=3$), Brackett ($n=4$), Pfund ($n=5$). The "first line" or "$H_\alpha$ line" is always from $n+1$ down to $n$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The wavelength of spectral lines is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.

Step 2: Formula Application:
For Paschen (1st line): $n_1 = 3, n_2 = 4$. $\frac{1}{\lambda_1} = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \frac{7}{144}$.
For Brackett (1st line): $n_1 = 4, n_2 = 5$. $\frac{1}{\lambda_2} = R \left( \frac{1}{16} - \frac{1}{25} \right) = R \frac{9}{400}$.

Step 3: Explanation:
$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{9/400}{7/144} = \frac{9}{400} \times \frac{144}{7} = \frac{9 \times 9}{25 \times 7} \times \text{simplified} = \frac{81}{175}$.

Step 4: Final Answer:
The ratio is 81 : 175.