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A photon of wavelength 0.1 A is emitted ...

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

A

0.05eV

B

1.05 eV

C

2.05 eV

D

3.05 eV

Text Solution

Verified by Experts

The correct Answer is:
C
`E=(p^2)/(2m)=(h^2)/(2mlamda^2)`
`=((6.62xx10^(-34))^2)/(2xx4xx1.67xx10^(-27)xx(0.1xx10^(-10))^2)`
`xx(1)/(1.6xx10^(-19))eV`
`=(43.82xx10^(-68))/(21.376xx10^(-68))=2.05eV`
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