Question

If an electron in hydrogen atom jumps from an orbit of level $n=3$ to an orbit of level $n=2$, emitted radiation has a frequency(R$=$ Rydberg's constant, C$=$ velocity of light).

• A. $\frac{3RC}{27}$
• B. $\frac{RC}{25}$
• C. $\frac{8RC}{9}$
• D. $\frac{5RC}{36}$

Answer 1

Answer: (D)

$\frac{5RC}{36}$

Given : $n_1=2$ and $n_2=3$

Wavelength emitted due to transition from $n_2$ to $n_1$ $\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\therefore$ $\frac{1}{\lambda }=R[\frac{1}{2^2}-\frac{1}{3^2}]=\frac{5R}{36}$

Thus frequency of radiation emitted $\nu =\frac{C}{\lambda }$

$⟹$ $\nu =\frac{5RC}{36}$