JEE PYQ: Atomic Physics Question 21
Question 21 - 2021 (26 Feb Shift 1)
If $\lambda_1$ and $\lambda_2$ are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of $\lambda_1 : \lambda_2$ is:
(1) 1 : 3
(2) 1 : 9
(3) 7 : 135
(4) 7 : 108
Show Answer
Answer: (3)
Solution
For Lyman series (3rd member): $n_1 = 1, n_2 = 4$. $\frac{1}{\lambda_1} = R\left(1 - \frac{1}{16}\right) = \frac{15R}{16}$, so $\lambda_1 = \frac{16}{15R}$. For Paschen series (1st member): $n_1 = 3, n_2 = 4$. $\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$, so $\lambda_2 = \frac{144}{7R} = \frac{9 \times 16}{7R}$. $\frac{\lambda_1}{\lambda_2} = \frac{16}{15R} \times \frac{7R}{9 \times 16} = \frac{7}{135}$.
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