Question 6

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

= -13.6/1^{2} = -13.6 eV

The atom is excited to a higher level, n^{2} = 4.

Let E^{2} 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 = 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 × 10^{8} m/s

∴ *λ = hc/E*

= (6.6x10^{-34}x3x10^{8})/(2.04x10^{-18})

= 9.7x10^{-8} m = 97 nm

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

*v = c/λ*

= (3x10^{8})/(9.7x10^{-8}) ≈ 3.1 x 10^{15} Hz

Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 10^{15} Hz.

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