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Question
A beam of light having wavelengths distributed uniformly between 450 nm to 550 nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam?
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Solution
Given:
Minimum wavelength of the light component present in the beam, λ1 = 450 nm Energy associated (E1) with wavelength (λ1 ) is given by
E1 = `(hc)/(lamda_1)`
Here,
c = Speed of light
h = Planck's constant
`therefore E_1 = (1242)/(450)`
= 2.76 eV
Maximum wavelength of the light component present in the beam, = 550 nm
Energy associated (E2) with wavelength is given by
`E_2 = (hc)/( lamda)`
∴ `E_2 = 1242/550 = 2.228 = 2.26 eV`
The given range of wavelengths lies in the visible range.
`therefore n_1 = 2, n_2 = 3,4,5....`
Let E'2 , E'3 , E'4 and E'5 be the energies of the 2nd, 3rd, 4th and 5th states, res pectively.
`E_2 - E_3 = 13.6(1/4 - 1/9)`
=`(12. 6 xx 5)/30=1.9 eV`
`E_2- E_4 = 13.6 (1/4 - 1/16)`
= 2.55 eV
`E_2-E_2 = 13.6 (1/4 - 1/25)`
= `(10.5xx21)/100 = 2.856 eV`
Only, E'2 − E'4 comes in the range of the energy provided. So the wavelength of light having 2.55 eV will be absorbed.
`lamda = (1242)/2.55 = 487.5 nm`
= 487 nm
The wavelength 487 nm will be absorbed by hydrogen gas. So, wavelength 487 nm will have less intensity in the transmitted beam.
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