Question:
An electron in a hydrogen-like atom is in an excited state. It has total energy of -3.4 eV. The kinetic energy and de-Broglie wavelength respectively are:
An electron in a hydrogen-like atom is in an excited state. It has total energy of -3.4 eV. The kinetic energy and de-Broglie wavelength respectively are:
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For hydrogen atom, kinetic energy = magnitude of total energy.
Updated On: Mar 18, 2026
- 3.4 eV, 0.66 nm
- -3.4 eV, 1.99 nm
- 2.8 eV, 2.38 nm
- 1.1 eV, 1.28 nm
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The Correct Option is A
Solution and Explanation
Step 1: Relation of energies. E = -K K = 3.4 eV
Step 2: de-Broglie wavelength. λ = h√(2mK) ≈ 0.66 nm
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