⚛️ Physics · Atoms · NEET & JEE

A hydrogen atom emits radiation when an electron falls from n = 4 to n = 2. What is the energy of the emitted photon?

Answer: 2.55 eV.

A 1.51 eV
B 2.55 eV
C 3.40 eV
D 10.20 eV

Correct answer: B. 2.55 eV

Explanation: The emitted photon energy is 13.6(1/2<sup>2</sup> - 1/4<sup>2</sup>) eV = 13.6(3/16) = 2.55 eV.

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.

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A hydrogen atom emits radiation when an electron falls from n = 4 to n = 2. What is the energy of the emitted photon?

Concept context

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