Question

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

• A. $\left(16/25\right){\lambda }_0$
• B. $\left(20/27\right){\lambda }_0$
• C. $\left(27/20\right){\lambda }_0$
• D. $\left(25/16\right){\lambda }_0$

Answer 1

The correct option is B $\left(20/27\right){\lambda }_0$

As in both cases electron came to second orbit, we will consider Balmer series,

Now, for Balmer series, wavelength is given by,

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2}),n=3,4,5$

In case1, when n=3, $\lambda ={\lambda }_0$

$\frac{1}{{\lambda }_0}=R(\frac{1}{4}-\frac{1}{9}),n=3,4,5R=\frac{36}{5{\lambda }_0}\to \left(1\right)$

In case2, when n=4, $\lambda$ is given by

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{4^2})\frac{1}{\lambda }=R(\frac{1}{4}-\frac{1}{16})=\frac{27}{20{\lambda }_0}\lambda =\frac{20{\lambda }_0}{27}$