JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

The energy of an electron in an orbit of the Bohr's atom is $- 0.04 E_{0}$ eV where $E_{0}$ is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then $\frac{2 π L}{h}$ is __________ :

Solution

Bohr's quantisation: $L = \frac{nh}{2 π}$ , so $\frac{2 π L}{h} = n$ — exactly the principal quantum number.

Energy in the $n$ -th orbit: $E_{n} = - \frac{E _{0}}{n ^{2}}$ . Given $E_{n} = - 0.04 E_{0}$ , we get $n^{2} = \frac{1}{0.04} = 25$ , so $n = 5$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →