Question

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. What is the value of n?

(A) 3
(B) 4
(C) 5
(D) 2

Answer 1

The value of 'n' for which an electron in the hydrogen atom jumps to the ground state, resulting in a wavelength that illuminates a photosensitive material with a work function of 2.75 eV, leading to the correct option (C).

Step-by-step explanation:

When an electron falls back to the ground state (n=1) from an excited state 'n', it emits a photon of a specific wavelength. The energy of the emitted photon can be calculated using the Rydberg formula for hydrogen: E = 13.6eV * (1 - 1/n^2)

This energy becomes the kinetic energy of the ejected photoelectron from the photosensitive surface, but you must first subtract the work function (φ) of the surface: Kinetic Energy (K.E.) = Photon Energy - Work Function. The potential energy of the photoelectron in the stopping potential (V_s) is given by: K.E. = e * V_s.

Using the given work function φ = 2.75 eV and a stopping potential V_s = 10 V, we can write: 13.6eV * (1 - 1/n^2) - 2.75eV = e * 10 V Solve for 'n', keeping in mind that 'e' is 1.6 x 10^-19 Coulombs: 13.6eV * (1 - 1/n^2) - 2.75eV = 1.6 x 10^-19 C * 10 V. Through calculations, we can find that the 'n' value corresponding to the correct energy is n=5, which matches option (C).