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 is $\rho \pi a_0$ (where a0 is the Bohr radius). The value of $\rho$ is

Answer 1

$mvr=\frac{3h}{2\pi }$
$\Rightarrow \lambda =\frac{2\pi r}{3}$
Also, $\lambda =\frac{2\pi r}{3}=\frac{2\pi a_0{n}^{2}lZ}{3}(Q_r=\frac{a_0{n}^2}{Z})$
$=\frac{2\pi }{3}\times a_0\left(\frac{9}{3}\right)$
$=2\pi a_0$
$\Rightarrow \rho =2$