Question

# The wavelength of the emitted radiation, if electron in hydrogen atom jumps from the third orbit to second orbit is

A
λ=365R
B
λ=5R36
C
λ=5R
D
λ=R6

Solution

## The correct option is A $$\lambda =\dfrac{36}{5R}$$Wave number, $$\bar{v}=R\left [ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right ]$$$$\Rightarrow \lambda =\dfrac{1}{R[\dfrac{1}{4}-\dfrac{1}{9}]}$$$$\Rightarrow \lambda =\dfrac{36}{5R}$$Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More