JEE PYQ: Atomic Physics Question 56
Question 56 - 2019 (12 Apr Shift 2)
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, $\lambda_1/\lambda_2$, of the photons emitted in this process is:
(1) 20/7
(2) 27/5
(3) 7/5
(4) 9/7
Show Answer
Answer: (1)
Solution
$\frac{1}{\lambda_1} = \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = \frac{7R}{16 \times 9}$. And $\frac{1}{\lambda_2} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = \frac{5R}{36}$. Now $\frac{\lambda_1}{\lambda_2} = \frac{(5R/36)}{7R/(16 \times 9)} = \frac{20}{7}$.
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