Question

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

Wavelength of radiation in hydrogen atom is given by
$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]\Rightarrow \frac{1}{{\lambda }_0}=R[\frac{1}{2^2}-\frac{1}{3^2}]=R[\frac{1}{4}-\frac{1}{9}]=\frac{5}{36}R$ ...... (i)
and $\frac{1}{{\lambda }^{{}^{\prime }}}=R[\frac{1}{2^2}-\frac{1}{4^2}]=R[\frac{1}{4}-\frac{1}{16}]=\frac{3R}{16}$ ....... (ii)
From equation (i) and (ii) $\frac{{\lambda }^{{}^{\prime }}}{\lambda }=\frac{5R}{36}\times \frac{16}{3R}=\frac{20}{27}\Rightarrow {\lambda }^{{}^{\prime }}=\frac{20}{27}{\lambda }_0$