Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# The first member of Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is

Solution:

$$\begin{array}{l}\Delta E = \frac{hC}{\lambda } = \Delta E_{o}\left [ \frac{1}{4} – \frac{1}{9} \right ]\end{array}$$
….i

$$\begin{array}{l}So, \frac{hC}{\lambda’} = \Delta E_{o}\left [ \frac{1}{4} – \frac{1}{16} \right ]\end{array}$$
….ii

$$\begin{array}{l}\Rightarrow \frac{\lambda’}{\lambda} = \frac{5\times 16}{9\times4 \times3 }\end{array}$$

$$\begin{array}{l}\Rightarrow \lambda’ = \frac{5\times4\times 656.1}{9\times3 } = 486\: nm\end{array}$$