Calculate the energy (in joules) of a photon emitted when an electron in a hydrogen atom undergoes a transition from 𝑛=6 to 𝑛=1.

Answer 1

Use ∆E = R_H (1/n_f2 - 1/n_i2)

∆E = change in energy = ?

R_H = -2.179x10^-18 J

n_f = final energy level = 1

n_i = initial energy level = 6

∆E = -2.179x10^-18 J (1/1 - 1/36) = -2.179x10^-18 J (0.9722)

∆E = -2.12x10^-18 J

(Note the negative value of ∆E. This is because energy is emitted as an electron falls from a higher energy level to a lower energy level)