Question

A photon is emitted from a hydrogen atom when an electron goes from the n = 6 to the n = 2 state. What kind of electromagnetic radiation is this? Select one:

a. Visible Light

b. Ultraviolet Light

c. Infrared Radiation

d. X-Ray Radiation

e. Microwave Radiation

Answer 1

Solution:   When an elctron jump from n_i =6 to n_f =2

Then for the transition for ni=6 to nf =2, by using the Rydberg equation.

1/ λ=R x (1/nf²−1/ni²)

where λ is the wavelength of the emittted photon

and R is the Rydberg constant, which is equal to 1.097x10⁷ m⁻¹

On putting the values,

1/λ=1.097x10⁷ m⁻¹x (1/2²−1/6²)

1/ λ=2.4378x10⁶ m⁻¹ hence,

λ=4.10 x10⁻⁷m = 410 nm

So, you know that when an electron falls from ni=6 to nf=2, a photon of wavelength 410 nm is emitted and this range of wavelength is shown by Visible light (400- 700 nm).