In a process a neutron which is initially at rest, decays into a proton, an electron and an antineutrino. The ejected electron has a momentum of p 1 = 2.4 × 10 –26 kg-m/s and the antineutrino has p 2 = 7.0 × 10 –27 kg-m/s. Find the recoil speed of the proton if the electron and the antineutrino are ejected
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
= 18.6 m/s
= 15.0 m/sec
P 1 = 2.4 × 10 –26 kg–m/sec.
P 2 = 7.0 × 10 –27 kg–m/sec
P 1 + P 2 + P 3 = 0 ∴ P 3 = – (24 ×10 –27 + 7.0 × 10 –27 )
P 3 = 31 × 10 –27 ∴ V 3 = 
= 18.6 m/sec.
P e = 2.4 × 10 –26 
an = 7 0 × 10 –27 
p = –( 
e + 
an ) = – (24 × 10 –27 
+ 7.0 × 10 –27 
)
= 
×10 –27
V p = 
= 15.0 m/sec.
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