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, then the value of n is (1) 5 (2) 2 (3) 3 (4) 4

Answer 1

Explanation:

Given 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 photo electron is 10 v, then the value of n is

• Now the electron jumps from an exited state to ground state.
• So in the ground state n = 1. Now there is a wavelength λ and the work function is given by φ = 2.75 eV
• Stopping potential Vo = 10 V
• The energy released E = work function + kinetic energy
• So E = K.E + φ
• E = e Vo + φ
•     = e x 10 + 2.75 e
• Energy E = 12.75 eV
• Now we need to find the value of n
• Now E = - 13.6 Z^2 / n^2 is in the ground state.
• So E1 = - 13.6 eV
• So if we put n = 4 and z = 1 we get
•    E2 = - 13.6 x 1 / 16
•           = - 0.85
• Now E2 – E1 will be
•    So – 0.85 – (- 13.6) = 12.75 eV
• So both the energy will be equal when n = 4

Reference link will be

https://brainly.in/question/16271581