A photon of energy 10.2 eV corresponds to light of wavelength $λ_{0} .$ 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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$λ = λ_{0}$

$λ > λ_{0}$

Insufficient data to reach a conclusion

$λ < λ_{0}$

answer is C.

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Detailed Solution

Energy of photon + K.E. of atom = $13.6 (1 - \frac{1}{2^{2}}) e V = 10.2 e V$
$∴ \frac{hc}{λ} = \frac{hc}{λ_{0}} -$ KE of atom $\Rightarrow λ > λ_{0}$

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