Question

A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excited state, emitting a 466 A photon. Find a) The energy of the free electron, and b) The energy of the original photon

Answer 1

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Step-by-step explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

`h\\u_2 =(hc)/(\lambda)=(12400)/(466)`

= 26.6 e V

h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁ = 23.2 + 13.6

= 36.8 e V

energy of the original photon is 36.8 eV