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Question
An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 700 nm. Find the net energy absorbed by the atom in the process.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
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Solution
Given :-
Wavelength of absorbed photon, `λ_1` = 500 nm
Wavelength of emitted photon, `λ_2` = 700 nm
Speed of light, `c = 3 xx 10^8 "m/s"`
Planck's constant, h = `6.63 xx 10^-34 "Js"`
Energy of absorbed photon,
`E_1 = (hc)/λ_1 = (h xx 3 xx 10^8)/(500 xx 10^-9)`
Energy of emitted photon,
`E_2 = (hc)/λ_2 = (h xx 3 xx 10^8)/(700 xx 10^-9)`
Energy absorbed by the atom in the process :-
`E_1 - E_2 = hc[1/λ_1 - 1/λ_2]`
`= 6.63 xx 3[1/5 - 1/7] xx 10^-19`
`= 6.63 xx 3 xx 2/35 xx 10^-19`
`= 1.136 xx 10^-19 "J"`
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