Question

The wavelength of radiation is ${\lambda }_0$ When an electron jumps from third to second orbit of hydrogen atom. For the electron to 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 $\frac{20}{27}{\lambda }_0$

$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{2_2^2})$
$\frac{1}{{\lambda }_0}=R(\frac{1}{2^2}-\frac{1}{3^2})=R(\frac{1}{4}-\frac{1}{9})=\frac{5R}{36}$
$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{4^2})=R(\frac{1}{4}-\frac{1}{16})=\frac{3R}{16}$
$\frac{\lambda }{{\lambda }_0}=\frac{5}{36}\times \frac{16}{3}=\frac{20}{27}$