An electron in the hydrogen atom jumps from excited state n to the ground state

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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.

Related Questions:

1. The dimensions of mobility of charge carriers are
2. In a photoelectric effect experiment, the slope of the graph between the stopping
3. A ray of light travelling in a transparent medium of refractive index µ, falls on surface
4. The potential difference that must be applied to stop the fastest photoelectrons
5. In Young’s double slit experiment, the ratio of maximum and minimum intensities

Topics: Atoms and Nuclei (136)
Subject: Physics (2479)

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