The orbit angular momentum of e in nth shell of H atom is 3h/π . The total energy of e is

The correct option is –0.38 eV.

According to Bohr's quantization postulate, the orbital angular momentum ($L$) of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by the formula:

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

where:
• $n$ is the principal quantum number (shell number)
• $h$ is Planck's constant

We are given that the orbital angular momentum of the electron is:

$L=\frac{3h}{\pi }$

By equating the two expressions for the angular momentum:

$\frac{nh}{2\pi }=\frac{3h}{\pi }$

Cancelling the common term $\frac{h}{\pi }$ from both sides:

$\frac{n}{2}=3$

Multiplying both sides by 2 gives:

$n=6$

This indicates that the electron is in the 6^th shell.

The total energy ($E$) of an electron in the $n^{\text{th}}$ shell of a hydrogen atom is given by:

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

Substituting $n=6$ into the energy equation:

$E_6=-\frac{13.6}{6^2}\text{eV}$

$E_6=-\frac{13.6}{36}\text{eV}$

$E_6\approx -0.378\text{eV}\approx -0.38\text{eV}$

Thus, the total energy of the electron in this shell is –0.38 eV.