Question

If two deuterium nuclei get close enough together, the attraction of the strong nuclear force will fuse them to make an isotope of helium. The process will release a vast amount of energy. The range of nuclear force is ${10}^{-15}$ $m$. This is the principle behind the nuclear fusion reactor. The deuterium nuclei move much too fast to contain them by physical walls. Hence, they are confined magnetically How fast two nuclei shall come to have a head on collision to fuse?

• A. $1.2\times {10}^{-6}$ $m{s}^{-1}$
• B. $1.2\times {10}^7$ $m{s}^{-1}$
• C. $3.8\times {10}^6$ $m{s}^{-1}$
• D. $8.72\times {10}^6$ $m{s}^{-1}$

Answer 1

The correct option is B $1.2\times {10}^7$ $m{s}^{-1}$

Applying energy balance:
$KE$ = $\frac{q_1{q}_2}{4\pi {\epsilon }_0\left(r\right)}$
$\frac{m{v}^2}{2}$ = $\frac{q_1{q}_2}{4\pi {\epsilon }_{0}r}$
$V^2$$=\frac{1.6\times {10}^{-19}\times 1.6\times {10}^{-19}\times 9\times {10}^9\times 2}{{10}^{15}\times 2\times 1.67\times {10}^{-27}}$

$v$ $=1.2\times {10}^7$ $m{s}^{-1}$