Question

In order to fuse two nuclei, they must be brought at a separation of about 2 fm or less. Let two deuterium nuclei may be brought to fuse together by colliding them with equal and opposite velocities. Minimum speed required for above process must be around.

• A. $5.8\times {10}^{12}m{s}^{-1}$
• B. $5.8\times {10}^{10}m{s}^{-1}$
• C. $5.8\times {10}^{6}m{s}^{-1}$
• D. $5.8\times {10}^{4}m{s}^{-1}$

Answer 1

The correct option is C $5.8\times {10}^{6}m{s}^{-1}$

The complete process is a shown in the figure

Potential energy barrier for 2 fm separation $=U=\frac{k{q}_1{q}_2}{r}$
$=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{2\times {10}^{-15}}=1.15\times {10}^{-13}J$
On equating this with available KE, we get
$2\times \frac{1}{2}m{v}^2=1.15\times {10}^{-13}$
$\Rightarrow v^2=\frac{1.15\times {10}^{-13}}{2\times 1.67\times {10}^{-27}}$
$\Rightarrow v\approx 5.8\times {10}^{}m{s}^{-1}$