A photon of energy 10.2 eV corresponds to light of wavelength λ0. Due to an electron transition from n = 2 to n = 1 in a Hydrogen atom, light of wavelength λ is emitted. If we take into account the recoil of the atom when the photon is emitted, then

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answer is 3.

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Detailed Solution

Energy of photon + K.E. of atom = 13.6 1−122eV = 10.2eV∴ hcλ = hcλ0− KE of atom ⇒ λ > λ0

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