Question

Assume that two deutron nuclei in the core of fusion reactor at temperature $T$ are moving towards each other,each with kinetic energy $1.5kT$, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature $T$ required for them to reach a separation of $4\times {10}^{-15}$m is in the range

• A. $1.0\times {10}^{9}K$ < $T$<$2.0\times {10}^{9}K$
• B. $2.0\times {10}^{9}K$< $T$<$3.0\times {10}^{9}K$
• C. $3.0\times {10}^9$K < $T$<$4.0\times {10}^{9}K$
• D. $4.0\times {10}^{9}K$ < $T$<$5.0\times {10}^{9}K$

Answer 1

The correct option is A $1.0\times {10}^{9}K$ < $T$<$2.0\times {10}^{9}K$

Applying conservation of mechanical energy, we get:
Loss of kinetic energy of two deuteron nuclei $=$ Gain in their potential energy.
$\Rightarrow 2\times 1.5kT=\frac{1}{4\pi {\epsilon }_0}\frac{e\times e}{r}$
$\Rightarrow 2\times 1.5(8.6\times {10}^{-5}\frac{eV}{K})\times T=\frac{(1.44\times {10}^{-9}eVm)}{4\times {10}^{-15}m}$
$\Rightarrow T=\frac{1.44\times {10}^{-9}}{2\times 1.5\times 8.6\times {10}^{-15}\times 4\times {10}^{-15}}$$=0.0139\times {10}^{11}=1.4\times {10}^{9}K$