Question

An electron in an excited state of $L{i}^{2+}$ ion 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)

$L=\frac{nh}{2\pi }=\frac{3h}{2\pi }$
$n=3$
$\lambda =\frac{h}{mv}=\frac{h\cdot r}{mvr}=\frac{hr}{\frac{3h}{2\pi }}$
$=\frac{2}{3}\pi r$
For $L{i}^{2+}$ atom radius of orbit.
$r=r_0\frac{n^2}{z}$
$r=a_0\times \frac{3^2}{3}$
$(a_0=$ Bohr radius)
$\therefore \lambda =\frac{2}{3}.\pi \times a_0\times \frac{3^2}{3}$
$=2\pi a_0$
Comparing
$\therefore P=2$