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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
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A charged oil drop is suspended in a uniform field at 3×10⁴ V/m so that it neither falls
- 1 a.m.u. is equivalent to
- Which of the following phenomena does not show the wavenature of light?
- Two capacitors 3μF and 4μF, are individually charged across a 6 V battery.
- The dimension of 1/2ε₀E², where ε₀ is permittivity of free space and E is electric field
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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