Question

An electron is moving in $2^{nd}$ excited orbit of H-atom Radius of orbit in terms of de-Broglie wavelength $\lambda$ of electron can be given as

• A. $\frac{\lambda }{\pi }$
• B. $\frac{2\lambda }{\pi }$
• C. $\frac{3\lambda }{2\pi }$
• D. $\frac{\lambda }{2\pi }$

Answer 1

The correct option is C $\frac{3\lambda }{2\pi }$

From Bohr's postulate,

$mvr=\frac{nh}{2\pi }$. . . . . .(1)

de-broglie wavelength, $\lambda =\frac{h}{mv}$

$mv=\frac{h}{\lambda }$. . . . . .(2)

Substitute equation (2) in equation (1), we get

$r\frac{h}{\lambda }=\frac{nh}{2\pi }$

$r=\frac{n\lambda }{2\pi }$

For 2nd excited state, $n=3$

Radius of orbit, $r=\frac{3\lambda }{2\pi }$

The correct option is C.