Q 12-12-102JEE MainJEE Main 2020 (8 Jan, Shift 2)Easy
The first member of the Balmer series of hydrogen atom has a wavelength of $6561$ Å. The wavelength of the second member of the Balmer series (in nm) is ______.
Numerical value type. Enter your answer.
Answer: 486
$\dfrac{1}{\lambda} = R\left(\dfrac{1}{2^2} - \dfrac{1}{n^2}\right)$.
First member ($n = 3$): $\dfrac{1}{4} - \dfrac{1}{9} = \dfrac{5}{36}$. Second member ($n = 4$): $\dfrac{1}{4} - \dfrac{1}{16} = \dfrac{3}{16}$.
$$\lambda_2 = \lambda_1\cdot\frac{5/36}{3/16} = 6561\times\frac{20}{27} = 4860\ \text{Å} = 486\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics