Question

In a hydrogen atom, an electron jumps from the first excited state to the ground state, and a photon is emitted. This photon is incident on a metal surface having a work function of 2eV. Calculate the stopping potential of the electron emitted from the metal surface.

Answer 1

Given: Initial state of electron in Hydrogen atom (n_i) = 2 (first excited state)

Final state of electron in Hydrogen atom (n_f) = 1 (ground state)

Work function of the metal surface (θ) = 2 eV

To find: The stopping potential (V_s) = ?

The energy of an electron in the nth energy level of a hydrogen atom is given by the formula:

E_n = `-13.6/n^2 eV`

Energy in the first excited state (n_i = 2):

E₂ = `-13.6/2^2 eV`

= `-13.6/4 eV`

= −3.4 eV

Energy in the ground state (n_f = 1):

E₁ = `-13.6/1^2 eV`

= −13.6 eV

The energy of the emitted photon during the transition is the difference between these two energy levels:

`E_"photon"` = E₂ − E₁

= (−3.4 eV) − (−13.6 eV)

= 10.2 eV

According to the photoelectric equation:

K_max = `E_"photon" - theta`

= 10.2 eV − 2 eV

= 8.2 eV

The maximum kinetic energy of the photoelectrons is also related to the stopping potential (V_s) by the formula:

K_max = e V_s

V_s = `K_"max"/e`

= `(8.2 eV)e`

= 8.2 V

∴ The stopping potential of the electron is 8.2 V.