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Carbon-II decays to boron-II according t...

Carbon-II decays to boron-II according to the following formula.
`._(6)^(11)Crarr ._(5)^(11)B+e^(+)+v_(e )+0.96 MeV`
Assume that positrons `(e^(+))` produced in the decay combine with free electrons in the atmosphere and annihilate each other almost immediately. Also assume that the neutrinos `(v_(e ))` are massless and do not interact with the environment. At t = 0 we have `1 mu g` of `._(6)^(12)C` . If the half-life of the decay process is `t_(0)`, the net energy produced between time t = 0 and `t=2t_(0)` will be nearly -

A

`8 xx 10^(18) MeV`

B

`8 xx 10^(16) MeV`

C

`4 xx 10^(18) MeV`

D

`4 xx 10^(16) MeV`

Text Solution

Verified by Experts

The correct Answer is:
B
Amoutn decay `= 0.75 xx 10^(-6)` gm
Number of atoms `= 0.4 xx 10^(17)`
Energy (in nuclear reaction) `= 0.96 xx` number of atoms `~~4 xx 10^(16)MeV`
By `E= mc^(2)`
As 1 electron from reaction annihilate by another electrons from atmosphere so energy from annihilation `=2mc^(2) ~~ 4 xx 10^(16) MeV`. Total energy `=8 xx 10^(16) MeV`
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