The energy of an electron in an orbit in hydrogen atom is 3-4 eV. Its angular momentum in the orbit will be :

The energy of an electron in an orbit in hydrogen atom is 3-4 eV. Its angular momentum in the orbit will be :
(A) (h/2π
(B) 2h/π
(C) h/π
(D) h/2π

Answer: (C) h/π

Step 1: Recall formulas for hydrogen atom (Bohr model)

1. Energy of electron in nth orbit:

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

where $n = 1,2,3,\dots$

2. Angular momentum in nth orbit:

$$L_n = n \frac{h}{2\pi}$$

Step 2: Find approximate $n$n from energy

We are given $E \approx 3\text{–}4 \text{ eV}$.$$E_n = -\frac{13.6}{n^2} \Rightarrow n^2 = \frac{13.6}{|E_n|}$$

Take the average $E \approx 3.4\ \text{eV}$:$$n^2 = \frac{13.6}{3.4} = 4 \Rightarrow n = 2$$

So the electron is in the second orbit.

Step 3: Compute angular momentum

$$L_n = n \frac{h}{2\pi} = 2 \frac{h}{2\pi} = \frac{h}{\pi}$$

Step 4: Answer

Angular momentum = h/π

Answer: (C) h/π