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Carbon-11 decays to boron-11 according to the following formula.
$$
{ }_{6}^{11} C \rightarrow{ }_{5}^{11} B+e^{+}+v_{e}+0.96 \mathrm{MeV}
$$
Assume that positrons $\left(\mathrm{e}^{+}\right)$produced in the decay combine with free electrons in the atmosphere and annihilate each other almost immediately. Also assume that the neutrinos $\left(\mathrm{v}_{\mathrm{e}}\right)$ are massless and do not intersect with the environment. At $\mathrm{t}=0$ we have 1 pg of ${ }_{6}^{12} C .$ If the half-life of the decay process is $t_{0}$, the net energy produced between time
$\mathrm{t}=0$ and $\mathrm{t}=2 \mathrm{t}_{0}$ will be nearly
Options:
$$
{ }_{6}^{11} C \rightarrow{ }_{5}^{11} B+e^{+}+v_{e}+0.96 \mathrm{MeV}
$$
Assume that positrons $\left(\mathrm{e}^{+}\right)$produced in the decay combine with free electrons in the atmosphere and annihilate each other almost immediately. Also assume that the neutrinos $\left(\mathrm{v}_{\mathrm{e}}\right)$ are massless and do not intersect with the environment. At $\mathrm{t}=0$ we have 1 pg of ${ }_{6}^{12} C .$ If the half-life of the decay process is $t_{0}$, the net energy produced between time
$\mathrm{t}=0$ and $\mathrm{t}=2 \mathrm{t}_{0}$ will be nearly
Solution:
1927 Upvotes
Verified Answer
The correct answer is:
$8 \times 10^{16} \mathrm{MeV}$
Amount decay $=0.75 \times 10^{-6} \mathrm{gm}$
Number of atoms $=0.4 \times 10^{17}$
Energy(in nuclear reaction) $=0.96 \times$ number of atoms
$\approx 4 \times 10^{16} \mathrm{MeV}$
By $\mathrm{E}=\mathrm{mc}^{2}$
As 1 electron from reaction annihilate by another electrons from atmosphere so
Energy from annihilation $=2 \mathrm{mc}^{2} \approx 4 \times 10^{16} \mathrm{MeV}$
Total energy $=8 \times 10^{16} \mathrm{MeV}$
Number of atoms $=0.4 \times 10^{17}$
Energy(in nuclear reaction) $=0.96 \times$ number of atoms
$\approx 4 \times 10^{16} \mathrm{MeV}$
By $\mathrm{E}=\mathrm{mc}^{2}$
As 1 electron from reaction annihilate by another electrons from atmosphere so
Energy from annihilation $=2 \mathrm{mc}^{2} \approx 4 \times 10^{16} \mathrm{MeV}$
Total energy $=8 \times 10^{16} \mathrm{MeV}$
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