Question

# Carbon-11 decays to Boron-11 according to the following formula.${}_6^{11}C \Rightarrow {}_5^{11}B + {e^ + } + {\upsilon _e} + 0.96MeV$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. $4 \times {10^{16}}MeV$

Verified
129k+ views
Hint: Calculate the remaining mass of carbon using $M = \dfrac{{{M_0}}}{{{2^n}}}$. Find the number of particles which will give the number of reactions and calculate energy of all particles.

Step by step solution:
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
$= 4 \times {10^{16}} + 4 \times {10^{16}}MeV \\$
$= 8 \times {10^{16}}MeV \\$

Hence option B is correct.

Note: In this type of question, for the conversion of units we need to revise all the relations between different quantities. Take care at the time of energy calculation, we have to calculate the number of atoms then multiply to energy releases by one atom.