Question

A photon with energy 2.24 eV is absorbed by a hydrogen atom.

(a) Find the minimum n for a hydrogen atom that can be ionized by such a photon.

(b) Find the speed of the electron released from the state in part (a) when it is far from the nucleus.

Answer 1

Given a photon with energy 2.24 eV is barely capable of causing

a photoelectric effect when it strikes a sodium plate.The photon is

absorbed by a hydrogen atom.

(a)For hydrogen atom

E_n = - (13.6/n²) eV,where n indicatesthe state.

Here photon energy is 2.24 eV.Therefore,if |E_n|< 2.24 eV then only hydrogen atom can be ionised.

For n = 1 the energy is

|E₁| = |- (13.6)/(1)²| = 13.6 eV

Therefore,the given photon cannot ionise the hydrogen atom asthe energy is larger than the incident photon energy.

For n = 2 the energy is

|E₂| = |- (13.6)/(2)²| = 3.4 eV

Therefore,the given photon cannot ionise the electron ofhydrogen atom in n = 2 state.

For n = 3 the energy is

|E₃| = |- (13.6)/(3)²| = 1.51 eV

Therefore,the given photon can ionise the electron of hydrogenatom in n = 3 state as the incident photon energy is greater thanthe electron energy in the n = 3 state.

(b)The remaining energy is converted into kinetic energy ofelectron.

Therefore,the kinetic energy of electron is

(1/2)m_ev² = (2.24 - 1.51) = 0.73 eV

where m_e is the mass of the electron and v is thevelocity of the electron.

Therefore,we get

v = [(2 x 0.73 x 1.6 x 10^-19)/(9.11 x10^-31)]^1/2

v = 5.06 x 10⁵ = 506x 10³ = 506 km/s.