12th Physics 2018 Set1 Delhi Board Paper Solution
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.
For ground level, n 1 = 1
Let E1 be the energy of this level. It is known that E1 is related with n1 as:
E1 = -13.6/n12 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.
∴ E2 = -13.6/n22 eV
= -13.6/42 = -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 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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