Discussion Forum

An electron in an excited state of Li²⁺ ion has angular momentum  $\frac{3h}{2\pi}$. 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 

Answer:

Explanation:

Angular momentum

                     $=n\left(\frac{h}{2\pi}\right)=3\left(\frac{h}{2\pi}\right)$

  $\therefore$   n=3

  Now,     $r_{n}\propto \frac{n^{2}}{z}$

 $\therefore$           $r_{3}= \frac{(3)^{2}}{3}(a_{0})=3a_{0}$

 Now,           $mv_{3}r_{3}=3\left( \frac{h}{2\pi}\right)$

$\therefore$   $mv_{3}(3a_{0})=3\left(\frac{h}{2\pi}\right)$

or            $\frac{h}{mv_{3}}=2\pi a_{0}$

or            $\frac{h}{P_{3}}=2\pi a_{0}$           (P=mv)

or      $\lambda_{3}=2\pi a_{0}$            $\lambda=\frac{h}{p}$

  $\therefore$     Answer is 2.