Question

What will be the angular momentum of an electron, if the energy of this electron in H-atom is $1.5eV$ (in J-s)?

• A. $1.05\times {10}^{-34}$
• B. $2.1\times {10}^{-34}$
• C. $3.15\times {10}^{-34}$
• D. $-2.1\times {10}^{-34}$

Answer 1

Answer: (C)

$3.15\times {10}^{-34}$

Solution:- (C) $3.15\times {10}^{-34}J/sec$

Energy of electron in $H$-atom $\left(E\right)=\frac{-13.6}{n^2}eV$

$\Rightarrow n^2=\frac{-13.6}{E}$

Given that energy of the electron in $H$-atom is $1.5eV$.

$\therefore n^2=\frac{-13.6}{-1.5}$

$\Rightarrow n=\sqrt{9.067}\approx 3$

As we know that, angular momentum $\left( p \right) is given by-

$p=\frac{nh}{2\pi }$

whereas, h is planck's constant $=6.6\times {10}^{-34}J/sec$

$\therefore p=\frac{3\times 6.6\times {10}^{-34}}{2\times 3.14}=3.15\times {10}^{-34}J/sec$

Hence the angular momentum of electron will be $3.15\times {10}^{-34}J/sec$.