Question

The wavelength of radiation emitted is ${\lambda }_0$ when an electron jumps from the third to the second orbit of hydrogen atom. For the electron Jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be

• A. $\frac{16}{25}{\lambda }_0$
• B. $\frac{20}{27}{\lambda }_0$
• C. $\frac{27}{20}{\lambda }_0$
• D. $\frac{25}{16}{\lambda }_0$

Answer 1

The correct option is B $\frac{20}{27}{\lambda }_0$

We have the equation

$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where R is Rydberg Constant

Then,

Transition from third orbit to second orbit

$\frac{1}{{\lambda }_0}=R(\frac{1}{2^2}-\frac{1}{3^2})=R(\frac{1}{4}-\frac{1}{9})=\frac{5R}{36}$

Transition from fourth to second orbit

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{4^2})=R(\frac{1}{4}-\frac{1}{16})=\frac{3R}{16}$

$\frac{\lambda }{{\lambda }_0}=\frac{5}{36}$ X $\frac{16}{3}=\frac{20}{27}$

So $\lambda =\frac{20}{27}{\lambda }_0$