Question

An electron in an excited state of $Li ^{2+}$ ions has angular momentum $\frac{3 h}{2 \pi}$. 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,
$L=\frac{n h}{2 \pi}=\frac{3 h}{2 \pi}$ ....(i)
$\therefore n=3$
$\lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{h r}{\frac{3 h}{2 \pi}}$ (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
$\lambda=\frac{2}{3} \pi a_0 \times 3=2 \pi a_0$
$\therefore P=2$