JEE PYQ: Atomic Physics Question 42
Question 42 - 2020 (08 Jan Shift 2)
The first member of the Balmer series of hydrogen atom has a wavelength of 6561 A. The wavelength of the second member of the Balmer series (in nm) is ____.
Show Answer
Answer: 486
Solution
$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. For 1st member of Balmer series: $n_1 = 2, n_2 = 3$. $\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$ …(i). For 2nd member: $n_1 = 2, n_2 = 4$. $\frac{1}{\lambda’} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$ …(ii). Dividing: $\frac{\lambda’}{\lambda} = \frac{5 \times 16}{9 \times 4 \times 3}$. $\lambda’ = \frac{5 \times 4 \times 656.1}{9 \times 3} = 486$ nm.
BETA
