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(a) Consider the D-T reaction (deuterium-tritium fusion)
$$
{ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \longrightarrow{ }_2^4 \mathrm{He}+\mathrm{n}
$$
Calculate the energy released in $\mathrm{MeV}$ in this reaction from the data:
$\mathrm{m}\left({ }_1^2 \mathrm{H}\right)=2.014102 \mathrm{u}$;
$\mathrm{m}\left({ }_1^3 \mathrm{H}\right)=3.016049 \mathrm{u}$
(b) Consider the radius of both deuterium and tritium to be approximately $2.0 \mathrm{fm}$. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction?
(Hint: Kinetic energy required for one fusion event = average thermal kinetic energy available with the interacting particles $=2(3 \mathrm{kT} / 2) ; \mathrm{k}=$ Boltzmann's constant, $\mathbf{T}=$ absolute temperature.)
$$
{ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \longrightarrow{ }_2^4 \mathrm{He}+\mathrm{n}
$$
Calculate the energy released in $\mathrm{MeV}$ in this reaction from the data:
$\mathrm{m}\left({ }_1^2 \mathrm{H}\right)=2.014102 \mathrm{u}$;
$\mathrm{m}\left({ }_1^3 \mathrm{H}\right)=3.016049 \mathrm{u}$
(b) Consider the radius of both deuterium and tritium to be approximately $2.0 \mathrm{fm}$. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction?
(Hint: Kinetic energy required for one fusion event = average thermal kinetic energy available with the interacting particles $=2(3 \mathrm{kT} / 2) ; \mathrm{k}=$ Boltzmann's constant, $\mathbf{T}=$ absolute temperature.)
Solution:
1832 Upvotes
Verified Answer
The $\mathrm{D}-$ T reaction is, ${ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \longrightarrow{ }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n}+\mathrm{Q}$.
(a) $\mathrm{Q}=\mathrm{m}\left({ }_1^2 \mathrm{H}\right)+\mathrm{m}\left({ }_1^3 \mathrm{H}\right)-\left[\mathrm{m}\left({ }_2^4 \mathrm{He}\right)+\mathrm{m}\left({ }_0^1 \mathrm{n}\right)\right] \times 931 \mathrm{MeV}$ $=(2.014102+3.016049-4.002603$
$-1.00867) \times 931=17.58 \mathrm{MeV}$.
(b) Coloumbic repulsion between 2 nuclei of ${ }_1^2 \mathrm{H}$ or ${ }_1^3 \mathrm{H}$ with $\mathrm{r}=2 \mathrm{fm}$ for both is,
$$
\begin{gathered}
=\frac{q^2}{4 \pi \epsilon_0(2 \mathrm{r})}=\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{2 \times 2 \times 10^{-15}} \mathrm{~J} \\
=\frac{9 \times 1.6 \times 1.6 \times 10^{+9-38+15}}{4}=5.76 \times 10^{-14} \mathrm{~J} .
\end{gathered}
$$
Kinetic energy required to overcome this repulsion is,
$\mathrm{K} . \mathrm{E}=\frac{3}{2} \mathrm{KT}$ where $\mathrm{T}=$ temperature
$\mathrm{K}=$ Boltzmann constant $=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$.
For 2 nuclei to interact,
$\mathrm{K} . \mathrm{E}=2 \times \frac{3}{2} \mathrm{KT}$ (2 moles interact)
$$
\begin{aligned}
\therefore \mathrm{T} &=\frac{\mathrm{KE}}{3 \mathrm{~K}}=\frac{5.76 \times 10^{-14} \mathrm{~J}}{3 \times 1.38 \times 10^{-23} \mathrm{JK}^{-1}} \\
&=1.39 \times 10^9 \mathrm{~K} .
\end{aligned}
$$
(a) $\mathrm{Q}=\mathrm{m}\left({ }_1^2 \mathrm{H}\right)+\mathrm{m}\left({ }_1^3 \mathrm{H}\right)-\left[\mathrm{m}\left({ }_2^4 \mathrm{He}\right)+\mathrm{m}\left({ }_0^1 \mathrm{n}\right)\right] \times 931 \mathrm{MeV}$ $=(2.014102+3.016049-4.002603$
$-1.00867) \times 931=17.58 \mathrm{MeV}$.
(b) Coloumbic repulsion between 2 nuclei of ${ }_1^2 \mathrm{H}$ or ${ }_1^3 \mathrm{H}$ with $\mathrm{r}=2 \mathrm{fm}$ for both is,
$$
\begin{gathered}
=\frac{q^2}{4 \pi \epsilon_0(2 \mathrm{r})}=\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{2 \times 2 \times 10^{-15}} \mathrm{~J} \\
=\frac{9 \times 1.6 \times 1.6 \times 10^{+9-38+15}}{4}=5.76 \times 10^{-14} \mathrm{~J} .
\end{gathered}
$$
Kinetic energy required to overcome this repulsion is,
$\mathrm{K} . \mathrm{E}=\frac{3}{2} \mathrm{KT}$ where $\mathrm{T}=$ temperature
$\mathrm{K}=$ Boltzmann constant $=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$.
For 2 nuclei to interact,
$\mathrm{K} . \mathrm{E}=2 \times \frac{3}{2} \mathrm{KT}$ (2 moles interact)
$$
\begin{aligned}
\therefore \mathrm{T} &=\frac{\mathrm{KE}}{3 \mathrm{~K}}=\frac{5.76 \times 10^{-14} \mathrm{~J}}{3 \times 1.38 \times 10^{-23} \mathrm{JK}^{-1}} \\
&=1.39 \times 10^9 \mathrm{~K} .
\end{aligned}
$$
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