Given the masses of various atomic particles mp=1.0072 u, mn=1.0087u, me=0.000548 u, mv¯=0, md=2.0141u, where P = portion, n = neutron, e = electron, v¯ = antineutrino and d = deuteron. Which of the following process is allowed by momentum and energy conservation?
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
n+n→ deuterium atom (electron bound to the nucleus)
b
p→n+e++v¯
c
e++e−→γ
d
n+p→d+γ
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
n+n→ deuterium atom is not energetically feasible as mass on product side is more than mass on reactant side.p→n+e++ν− is also not energetically feasible.e++e−→γ is not possible as momentum conservation does not hold.rather e++e−→2γ would be correct.As n+p→d+γ is energetically feasible as mass on product side is less than mass on reactant side, it could be possible.