Question

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

$L =\frac{ nh }{2 \pi}=\frac{3 h }{2 \pi} $
$n =3 $
$\lambda =\frac{ h }{ mv }=\frac{ h \cdot r }{ mvr }=\frac{ hr }{\frac{3 h }{2 \pi}}$
$=\frac{2}{3} \pi r$
For $Li ^{2+}$ 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
$\therefore P =2$