Question

Consider two deuterons moving towards each other with equal speeds in a deuteron gas.Their kinetic energies (when they are widely separated) shoud be such that the closest separation between them becomes 2 fm. Assume that the nuclear force is not effective for separations greater than 2 fm. At what temperature will the deuterons have this kinetic energy on an average?

• A. 500 K
• B. 10000 K
• C. ${10}^{5}K$
• D. ${10}^{9}K$

Answer 1

The correct option is D ${10}^{9}K$

Like we talked in the hint. We need to supply kinetic energy to the two nuclei to overcome the potential energy barrier.
That kinetic energy comes from heating and hence increasing temperature. But before that lets calculate how much potential energy is required to be increased. (We are doing this first so that we could supply exact amount of energy needed, nothing more, nothing less!!)
So, initially they are widely separated. So the distance between them is $\infty$.
$U_i=\frac{Q_1.Q_2}{4\pi {ϵ}_{0}r}$
Now, deuterons have one proton so charge $Q_1=Q_2=e$.
$r=\infty$
$Hence{U}_i=\frac{e^2}{4\pi {ϵ}_0\infty }=0$
Finally r =2fm
Hence
$U_f=\frac{e^2}{4\pi {ϵ}_0\left(2fm\right)}$
Now,
$\Delta U=\frac{e^2}{4\pi {ϵ}_0\left(2fm\right)}$
So, now if we provide kinetic energy to each nuclei, total kinetic energy will be 2K.
Hence, from our discussion
$2K=\frac{e^2}{4\pi {ϵ}_0\left(2fm\right)}$
$=\frac{(1.6\times {10}^{-19}C)\times (1.6\times {10}^{-19}C)\times (9\times {10}^{9}N{m}^2{C}^{-2})}{2\times {10}^{-15}m}$
Or,
$K=5.7\times {10}^{-14}J$.
Now, if the temperature of the gas is the average kinetic energy of random motion of each nucleus will be 1.5 kT.
The temperature needed for the deuterons to have the average kinetic energy of will be given by
$1.5kT=5.7\times {10}^{-14}J$.
Or,
$T=\frac{5.7\times {10}^{-14}}{1.5\times 1.38\times {10}^{-23}JK}$
$=2.8\times {10}^{9}K.$
Quite a high temperature. This is the reason these nuclear reactions are also called as Thermonuclear Fustion.