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In the decay .^(64)Cu rarr .^(64)Ni + e^...

In the decay `.^(64)Cu rarr .^(64)Ni + e^(-)+v_(1)` the maximum kinetic energy carried by the positron is found to be `0.680 MeV` (a) Find the energy of the neutrino which was emitted together with a positron of energy `0.180 MeV` (b) what is the momentum of this neutrion in `kg-m//s` ? Use the formula applicable to photon.

Text Solution

Verified by Experts

The correct Answer is:
(a) `(0.680-0.180)Me V =500 keV`
(b) `(500xx10^(3)e)/(c )=2.67xx10^(-22)kg-m//s`
(a) `KE "of"upsilon+KE "of"e^(+)= "Maximum energy of"e^(+)`
`rArrKE "of" upsilon=(0.680-0.180)MeV=500KeV`.
(b) `p=(E )/(C )=(500xx10^(3)xx1.6xx10^(-19))/(3xx10^(8))Kg-m//s=2.67xx10^(-22)Kg-m//s`.
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