Question

The wavelength of the radiation emitted when an electron falls from Bohr's orbit 4 to 2 in H atom is:

• A. 972 nm
• B. 486 nm
• C. 243 nm
• D. 182 nm

Answer 1

The correct option is A 972 nm

From Rydberg's equation,

$v=\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

For the second line of Balmer series of hydrogen spectrum, $n_1=2$ and $n_2=4.$

$v=R_H[\frac{1}{2^2}-\frac{1}{4^2}]$

$\frac{1}{\lambda }=R_H\left[\frac{3}{16}\right]\therefore \lambda =\frac{16}{3R}=972nm$

Hence, the correct option is $\text{A}$