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निम्न आकड़ों m(""(20)^(40)Ca)=39.962591u,...

निम्न आकड़ों `m(""_(20)^(40)Ca)=39.962591u,m(""_(2)^(41)Ca)=40.962278u,m_(n)=1.00865, 1u=931.5MeV//C^(2)` से न्यूट्रॉन पृथक ऊर्जा की गणना कीजिए

A

7.57 MeV

B

8.36 MeV

C

9.12 MeV

D

9.56 MeV

Text Solution

Verified by Experts

The correct Answer is:
B
`""_(20)^(41)Ca rarr ""_(20)^(40)Ca + ""_(0)^(1) n`
Mass defect `(Delta m) = [m (""_(20)^(40)Ca) + m_(n)- m (""_(20)^(41) Ca)]` = 0.008963u
`therefore` Neutron separation energy `= 0.008963 xx 931.5 MeV`
`~~ 8.36 MeV`
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