Modern Physics 4 Question 24
25. An electron in an excited state of $Li^{2+}$ ion has angular momentum $\frac{3 h}{2 \pi}$. The de Broglie wavelength of the electron in this state is $p \pi a _0$ (where $a _0$ is the Bohr radius). The value of $p$ is
(2015 Adv.)
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Solution:
- Angular momentum $=n \frac{h}{2 \pi}=3 \frac{h}{2 \pi}$
$$ \therefore \quad n=3 $$
Now, $r _n \propto \frac{n^{2}}{z}$
$\therefore \quad r _3=\frac{(3)^{2}}{3}\left(a _0\right)=3 a _0$
Now, $m v _3 r _3=3 \frac{h}{2 \pi}$
$\therefore \quad m v _3\left(3 a _0\right)=3 \frac{h}{2 \pi}$
or $\frac{h}{m v _3}=2 \pi a _0$ or $\quad \frac{h}{P _3}=2 \pi a _0 \quad(\because P=m v)$
or $\quad \lambda _3=2 \pi a _0$
$\lambda=\frac{h}{P}$
$\therefore \quad$ Answer is 2 .
