Question

Angular momentum of an electron in bohr's hydrogen atom whose energy is -3.4 ev

Answer 1

Since it's having energy corresponding to 3.4 eV, therefore the electron lies in 2nd shell that is, n=2

So, angular momentum, mvr =nh/2pi = 2h/2pi = h/pi.

Answer 2

Answer : The angular momentum of an electron in bohr's hydrogen atom is $\frac{h}{\pi}$

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-\frac{13.6}{n^2}$

where,

n = number of orbit of hydrogen atom

$E_n$ = energy = -3.4 V

Now put all the given values in the above formula, we get:

$-3.4=-\frac{13.6}{n^2}$

$n=2$

Now we have to calculate the angular momentum of an electron in bohr's hydrogen atom.

Formula used :

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

So,

$L=\frac{2h}{2\pi}$

$L=\frac{h}{\pi}$

Therefore, the angular momentum of an electron in bohr's hydrogen atom is $\frac{h}{\pi}$