Question

The wavelength of radiation emitted is ${\lambda }_0$, when an electron jumps from the third to second orbit of a $H^{+}$ atom. For and electron jumping from fourth to second orbit, what would be the wavelength of the emitted radiation ?

• A. $\frac{16}{25}{\lambda }_0$
• B. $\frac{20}{27}{\lambda }_0$
• C. $\frac{27}{20}{\lambda }_0$
• D. $\frac{9}{25}{\lambda }_0$

Answer 1

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

Wavelength of emitted radiation is given by,

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

Transition from 3rd orlit to 2nd orlit
$\frac{1}{{\lambda }_0}=R[\frac{1}{2^2}-\frac{1}{3^2}]=R\left[\frac{5}{36}\right]$

Transition from 4th to 2 nd orlit
$\frac{1}{\lambda }=R[\frac{1}{2^2}-\frac{1}{4^2}]=R\left[\frac{3}{16}\right]$

Division of both equations gives,

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

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

Hence, $\left(B\right)$ is the correct answer.