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An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, the value of n is
Options
(a) 3
(b) 4
(c) 5
(d) 2
Correct Answer:
4
Explanation:
KEₘₐₓ = 10 eV ? = 2.75 eV
Total incident energy
E = ? + KEₘₐₓ = 12.75 eV
Energy is released when electron jumps from the excited state n to the ground state.
E₄ – E₁ = {-0.85 – (-13.6) eV}
= 12.75 eV value of n = 4.
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Topics: Atoms and Nuclei
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Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Photoelectric current from a given photocell will depend upon
- A planet moving along an elliptical orbit is closest to the sun at a distance of r₁
- An observer moving away from a stationary source. The observed frequency is half
- An electron in hydrogen atom makes a transition n₁ → n₂
- A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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