Question 18

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.

Answer

For ground level, n _{1 }= 1

Let E_{1} be the energy of this level. It is known that E_{1} is related with n_{1} as:

E_{1} = -13.6/n_{1}2 eV

= -13.6/12 = -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

= -13.6/42 = -13.6/16 eV

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

E = E_{2} - E_{1}

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

= 13.6 X 15/16 eV

= (13.6 X 15/16) X 1.6 X 10-19 = 2.04 X 10-18 J

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

E = hc/λ

Where,

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

c = Speed of light = 3 × 108 m/s

∴ λ = hc/E

= (6.6x10-34x3x108)/(2.04x10-18)

= 9.7x10-8 m = 97 nm

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

v = c/λ

= (3x108)/(9.7x10-8) ≈ 3.1 x 1015 Hz

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

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