STANDARD Physics Energy Levels MCQ Question
An electron is in an excited state in a hydrogen like atom. It has a total energy of –3.4 eV. The kinetic energy of the electron is E and its de-Broglie wavelength is λ. Then
E = 6.8 eV, λ = 6.6 × 10⁻¹⁰ m
E = 3.4 eV, λ = 6.6 × 10⁻¹⁰ m
E = 3.4 eV, λ = 6.6 × 10⁻¹¹ m
E = 6.8 eV, λ = 6.6 × 10⁻¹¹ m
Correct Answer
Detailed Explanation
The total energy is twice the kinetic energy with opposite sign. Thus, E = 3.4 eV. Using the de-Broglie wavelength formula, λ = h/√(2mE), we find λ = 6.6 × 10⁻¹⁰ m.
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