JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

The energy of an electron in the n^ th stationary state of a one-electron system is given by the Bohr's formula: E_n = -2.18 10^ -18 Z^2 n^2 J where Z is the atomic number and n is the principal quantum number. For option (1): Second orbit of He^+ (Z=2, n=2) E = -2.18 10^ -18 2^2 2^2 = -2.18 10^ -18 J. (Incorrect sign in option) For option (2): Third orbit of Li^ 2+ (Z=3, n=3) E = -2.18 10^ -18 3^2 3^2 = -2.18 10^ -18 J. (Correct) For option (3): Second orbit of H atom (Z=1, n=2) E = -2.18 10^ -18 1^2 2^2 = -0.545 10^ -18 J. (Incorrect value) For option (4): First orbit of He^+ (Z=2, n=1) E = -2.18 10^ -18 2^2 1^2 = -8.72 10^ -18 J. (Incorrect sign in option)