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A free atom of iron emits Kalpha X-rays ...

A free atom of iron emits `K_alpha` X-rays of energy 6.4 keV. Calculate the recoil kinetic energy of the atom. Mass of and iron atom `= 9.3 xx 10 ^(-26) kg`.

A

`39 times 10^(-4)eV`

B

`3.9 times 10^(-4)eV`

C

`0.39 times 10^(-4) eV`

D

`3.9 times 10^(-2)eV`

Text Solution

Verified by Experts

The correct Answer is:
B
Momentum of recoiling atom = momentum of the photon `=E/c`
K.E of recoiling atom `=p^(2)/(2m)=E^(2)/(2mc^(2))`
`" "=((6.4 times 10^(3) times 1.6 times 10^(-19))/(3 times 10^(8)))^(2) times 1/(2 times (9.3 times 10^(-26)))J`
`" "=0.63 times 10^(-22)J=3.9 times 10^(-4)eV`
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