JEE PYQ: Atomic Structure Question 20
Question 20 - 2020 (04 Sep Shift 2)
The shortest wavelength of H atom in the Lyman series is $\lambda_1$. The longest wavelength in the Balmer series of He$^+$ is:
(1). ($\frac{36\lambda_1}{5}$)
(2). ($\frac{5\lambda_1}{9}$)
(3). ($\frac{9\lambda_1}{5}$)
(4). ($\frac{27\lambda_1}{5}$)
Show Answer
Answer: (3)
Solution
$\frac{1}{\lambda_1} = R_H$. For He$^+$ Balmer ($3 \to 2$): $\frac{1}{\lambda} = \frac{5R_H}{9}$. So $\lambda = \frac{9\lambda_1}{5}$.
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