A free neutron at rest, decays into three particles: a proton, an electron and an anti-neutrino. 01n→11P+−10e+V¯In a particular decay, the antineutrino was found to have a total energy (including rest mass energy) of 0.0004 MeV and the momentum of proton was found to be equal to the momentum of electron. The kinetic energy of the electron (in MeV)Take mass of neutron=1.008658u , mass of proton =1.007u , mass of electron =0.000548u
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answer is 0.78.
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Detailed Solution
From conservation of energy mnc2=mpc2+kp+mec2+ke+mvc2+kv939.5656=938.2723+0.5109+0.0004+kp+ke∵mvc2+kv=0.0004MeV⇒ kp+ke=0.782MeV⇒ P22mP+P22me=0.782MeV⇒ P22me1+memp=0.782⇒ke=mpmp+me×0.7820≃0.7820MeV