Question

Carbon-11 decays to boron-11 according to the following formula. ${}_6^{11}C{\to }_5^{11}B+e^{+}+v_e+0.96MeV$ 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. $4\times {10}^{18}MeV$
• 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 is annihilated by another electron from atmosphere so Energy from annihilation $=2m{c}^2\approx 4\times {10}^{16}MeV$
Total energy $=8\times {10}^{16}MeV$