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

Answer 1

Answer: (B)

$\frac{\lambda }{{\lambda }_0}=\frac{\left[\frac{1}{2^2}-\frac{1}{3^2}\right]}{\left[\frac{1}{2^2}-\frac{1}{4^2}\right]}$
$=\frac{5}{36}\times \frac{16}{3}$
$=\frac{20}{27}$ or
$\lambda =\frac{20}{27}{\lambda }_0$