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Q.

A photon of wavelength 0.1 Ao is emitted by a helium atom as a consequence of the emission of photon. The KE gained by helium atom is

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a

0.05 eV

b

1.05 eV

c

2.05 eV

d

3.05 eV

answer is C.

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Detailed Solution

E=p22m=h22mλ2=6.62×10−3422×4×1.67×10−27×0.1×10−102×11.6×10−19eV=43.82×10−6821.376×10−68=2.05eV
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