Question

An electron in an excited state of $L{i}^{2+}$ ions has angular momentum $\frac{3h}{2\pi }$. The de Broglie wavelength of electron in this state is $P\pi a_0$ (where $a_0=$ Bohr radius). The value of $P$ is

• A. 3
• B. 2
• C. 1
• D. 4

Answer 1

Answer: (B)

Angular momentum,
$L=\frac{nh}{2\pi }=\frac{3h}{2\pi }$ ....(i)
$\therefore n=3$
$\lambda =\frac{h}{mv}=\frac{hr}{mvr}=\frac{hr}{\frac{3h}{2\pi }}$ (from (i)
$\therefore \lambda =\frac{2}{3}\pi r$ ....(ii)
For $L{i}^{2+}$, Radius of orbit, $r=a_0\frac{n^2}{Z}$
($a_0$ Bohr radius )
$\therefore r=a_0\times \frac{3^2}{3}$
Putting this value of $r$ in eq (ii), we get
$\lambda =\frac{2}{3}\pi a_0\times 3=2\pi a_0$
$\therefore P=2$