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${}_6^{11}C \Rightarrow {}_5^{11}B + {e^ + } + {\upsilon _e} + 0.96MeV$

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 (${\upsilon _e}$) are massless and do not intersect with the environment. At $t=0$ we have 1 $ \mu g$ of ${}_6^{12}C $. If the half time of the decay process is ${t_0}$, then what will be the net energy produced between time $t=0$ and $t=2{t_0}$ ?

A. $8 \times {10^{18}}MeV$

B. $8 \times {10^{16}}MeV$

C. $4 \times {10^8}MeV$

D. $4 \times {10^{16}}MeV$

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1. At t=0 mass is 1$\mu g$ I.e., ${M_0} = 1\mu g$ where ${M_0}$ is the rest mass of the particle.

${t_{\dfrac{1}{2}}} = {t_0}$

Now remained mass will be $M = \dfrac{{{M_0}}}{{{2^n}}}$ ( where $n = \dfrac{t}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{2{t_0}}}{{{t_0}}}$$ = 2$)

Therefore, $M = \dfrac{{{M_0}}}{{{2^2}}}$

$ \Rightarrow M = \dfrac{{1\mu g}}{4} = 0.25\mu g$ …….this is the amount of carbon remained

So used carbon will be =${M_0}$-M=$1\mu g - 0.25\mu g$

$ \Rightarrow 0.75\mu g = 0.75 \times {10^{ - 6}}g$

No. of moles of carbon =$\dfrac{{0.75 \times {{10}^{ - 6}}}}{{12}} = 0.0625 \times {10^{ - 6}}$ ($\because $1 mole of carbon contains 12 gram of carbon atoms)

That means this much amount of moles of carbon are used in the reaction.

No. of reactions will be=$0.0625 \times {10^{ - 6}} \times 6.023 \times {10^{23}}$ (where $6.023 \times {10^{23}}$ is Avogadro no.)

$ \Rightarrow 0.376 \times {10^{17}}$ reactions.

Now 0.96MeV energy is going into every reaction.

$\therefore $Total energy for the reaction is:

$

= 0.376 \times {10^{17}} \times 0.96MeV \\

= 0.36 \times {10^{17}}MeV \\

\approx 4 \times {10^{16}}MeV \\

$

As positrons produced in the decay combine with free electrons in the atmosphere and annihilate each other. Therefore, Annihilation energy for 1 reaction=$2{m_0}{c^2}$

$

\Rightarrow 2 \times 9.1 \times {10^{ - 31}} \times 9 \times {10^{16}} \\

\\

$

$ \Rightarrow 163.8 \times {10^{ - 15}}J$

Energy from annihilation for the reaction:

$

= 163.8 \times {10^{ - 15}} \times 0.376 \times {10^{17}}J \\

= \dfrac{{1.638 \times {{10}^{ - 15}}}}{{1.6 \times {{10}^{ - 19}}}} \times {10^{ - 6}} \times 0.376 \times {10^{17}}MeV \\

= 1.02 \times 0.376 \times {10^{17}}MeV \\

= 0.4 \times {10^{17}}MeV \\

= 4 \times {10^{16}}MeV \\

$

Now total Energy will be

$ = 4 \times {10^{16}} + 4 \times {10^{16}}MeV \\ $

$ = 8 \times {10^{16}}MeV \\$