Question

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.

A hydrogen atom initially in the ground state absorbs a photon, which excites it to the n = 4 level. Estimate the frequency of the photon.

Answer 1

For ground level, n₁ = 1

Let E₁ be the energy of this level. It is known that E₁ is related with n₁ as:

`E_1 = (-13.6)/(n_1^2) eV`

= `(-13.6)/1^2`

= −13.6 eV

The atom is excited to a higher level, n₂ = 4

Let E₂ be the energy of this level.

∴ `E_2 = (-13.6)/n_2^2 eV`

= `(-13.6)/4^2`

= `(-13.6)/16 eV`

The amount of energy absorbed by the photon is given as:

E = E₂ − E₁

= `(-13.6)/16 - ((-13.6)/1)`

= `(13.6 xx 15)/16 eV`

= `(13.6 xx 15)/16 xx 1.6 xx 10^(-19)`

= 2.04 × 10⁻¹⁸ J

For a photon of wavelength λ, the expression of energy is written as:

`"E" = (hc)/lambda`

Where,

h = Planck’s constant = 6.6 × 10⁻³⁴ Js

c = Speed of light = 3 × 10⁸ m/s

∴ λ = `(hc)/E`

= `(6.6 xx 10^-34 xx 3 xx 10^8)/(2.04 xx 10^-18)`

= 9.7 × 10⁻⁸ m

= 97 nm

And, frequency of a photon is given by the relation,

`v = c/lambda`

= `(3 xx 10^8)/(9.7 xx 10^(-8))`

≈ 3.1 × 10¹⁵ Hz

Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 10¹⁵ Hz.