⚛️ Physics · Atoms · NEET & JEE

An electron in hydrogen is initially in the n = 2 state. It absorbs a photon of energy 2.856 eV. Which final state is reached?

Answer: n = 5.

A n = 3
B n = 4
C n = 5
D n = 6

Correct answer: C. n = 5

Explanation: The energy difference from n = 2 to n = 5 is 13.6(1/4 - 1/25) = 2.856 eV. Therefore the absorbed photon raises the electron to n = 5.

Comparison of the Thomson model and the Rutherford model of the atom: in the Thomson plum pudding model alpha particles pass almost straight through a diffuse positive sphere, while in the Rutherford model a concentrated central nucleus deflects some alpha particles through large angles, with the lower panels showing the alpha particle source and gold foil and the observed result of wide-angle scattering

The Geiger–Marsden alpha-scattering result: a few alpha particles bounce back at large angles, which rules out Thomson’s diffuse model and demands a tiny, massive, positively charged nucleus. Image: Kurzon, CC BY 3.0, via Wikimedia Commons.

Concept context

Thomson's plum-pudding model, Rutherford's nuclear model from alpha-scattering, Bohr's postulates and hydrogen spectrum, spectral series (Lyman to Pfund), Rydberg formula, and limitations of each model.

Read the full Atoms notes →

An electron in hydrogen is initially in the n = 2 state. It absorbs a photon of energy 2.856 eV. Which final state is reached?

Concept context

More Atoms questions