Question

Calculate the energy of the emitted photon as well as the wavelength and frequency of electromagnetic radiation emitted from the hydrogen atom when the electron undergoes the transition from n = 5 to n = 4.

Answer 1

1) We know that 1/λ = R [ 1/n1^2- 1/n2^2]

λ = wavelength

R = Rydberg constant = 1.097 x 10⁷ m^-1

n1= 4, n2= 5

1/λ = R [ 1/n1^2- 1/n2^2]

substitute all the values

1/λ = 1.097 x 10⁷ m^-1. [ 1/4^2- 1/5^2]

1/λ = 1.097 x 10⁷ m^-1. [ 1/16 - 1/25 ]

On simplification,

λ = 40.5 x 10^-7 m

Therefore, wavelength λ   = 40.5 x 10^-7 m

2)

Energy E = hc/λ

               = (6.626 x 10^-34 J.s) (3 x 10⁸ m/s) / (40.5 x 10^-7 m)

               = 0.49 x 10^-19 J

           E = 0.49 x 10^-19 J

3)

Frequency v = E/h           [ since E = hv]

                   = 0.49 x 10^-19 J / 6.626 x 10^-34 J.s

                  = 0.074 x 10¹⁵ Hz

Hence,

frequency = 0.074 x 10¹⁵ Hz