Question

Carbon $-11$ decays to $boron-11$ according to the following formula.
${}_6^{11}C{\Rightarrow }_5^{11}B+e^{+}+v_e+0.96MeV$
Assume that positrons $\left(e^{+}\right)$ produced in the decay combine with free electrons in the atmosphere and annihilate each other almost immediately. Also assume that the neutrinos $\left(v_e\right)$ are massless and do not intersect 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. $8\times {10}^{18}MeV$
• B. $8\times {10}^{16}MeV$
• C. $4\times {10}^{8}MeV$
• D. $4\times {10}^{16}MeV$

Answer 1

The correct option is B $8\times {10}^{16}MeV$

Amount decay $=0.75\times {10}^{-6}gm$
Number of atoms $=0.4\times {10}^{17}$
Energy (in nuclear reaction) $=0,96\times$ number of atoms $MeV\approx 4\times {10}^{16}MeV$
By $E=m{c}^2$
As $1$ electron from reaction annihilate by another electrons from atmosphere so
Energy from annihilation $=2m{c}^2\approx 4\times {10}^{16}MeV$
Total energy $=8\times {10}^{16}MeV$.