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Calculate the kinetic energy of beta-par...

Calculate the kinetic energy of `beta`-particles and the radiation frequencies corresponding to the `gamma`-decays shown in figure.
Given, mass of `._12Mg^27` atom =26.991425 amu and mass of `._13Al^27` atom = 26.990080 amu

Text Solution

Verified by Experts

Radiatin frequency of `gamma _(1)` photon=
`v _(1) = (E _(3) - E _(2))/(h ) = ((1. 015 -0.834)MeV)/(6.62 xx 10 ^(-34) Js )`
`= (0.18 1 xx1.6 xx 10 ^(-13)J)/(6.62 xx 10 ^(-34)Js) = 4.37 xx 10 ^(19)s ^(-1)`
Radiation frequency of `gamma _(2)` photon =
`v _(2) = (E_(3) - E_(1))/(h)= ((1. 0.15 - 0) MeV)/( 6.62 xx 10 ^(-34) Js)`
` = 2.45 xx 10 ^(20) s ^(-1)`
Radiation frequency of `gamma _(3),` photon =
`v _(3) = (E _(2) - E_(1))/( h ) = (( 0.834 -0) MeV)/(6.62 xx 10 ^(-34)) =2.0 xx 10 ^(20) s ^(-1)`
Now emission of `beta _(1) ^(-)` particle is given by
` "" (12) M ^(27( to "" _(13) Al^(27) + beta ^(-) + v_(2) + Q`
` Q = [m ("" _(12) Mg ^(27)) - m ("" _(13) Al ^(27)) - E ( v _(2))`
`= [ 26. 991425-26.990080]u- (E_(3) - E_(1)) MeV`
`= 0.001345 xx 931 1. 015 MeV`
`therefore K.E of beta _(1) ^(-)` particla `= 0.23 6 MeV`
Emission of `beta _(2) ^(-)` particle is given by
` "" _(12)Mg ^(27) to "" _(13) Al ^(27) + beta _(2) ^(-) + v _(3) + Q`
`={[26.991425- 26.990080] 931- 0.834} MeV = 0.418 MeV`
`therefore K.E of beta _(2) ^(-)` particle `= 0.4 18 MeV`
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