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प्रश्न
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.
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उत्तर
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.
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