Question

An electron in an excited state of $L{i}^{2+}$ ion has angular momentum $\frac{3h}{2\pi }$. The de Broglie wavelength of the electron in this state in $p\pi {\alpha }_0$ (where ${\alpha }_0$ is the Bohr radius). The value of $p$ is

Answer 1

Angular momentum, $L=\frac{nh}{2\pi }=\frac{3h}{2\pi }$

$\Rightarrow n=3$

Also, Wavelength $\lambda =\frac{h}{p}=\frac{hr}{mvr}=\frac{hr}{\frac{3h}{2\pi }}=\frac{2\pi r}{3}$ ...(1)

But, $r=a_0\frac{n^2}{Z}$ ....(2)

Substituting (2) in (1), $\lambda =\frac{2\pi }{3}{a}_0\frac{n^2}{Z}=\frac{2\pi }{3}{a}_0\frac{3^2}{3}=2\pi a_0$

$\Rightarrow p=2$ comparing with the given expression $p\pi a_0$