QJEEAdv09Ph16b0: JEE Advanced 2009 Physics - Class 12 JEEAdv09Ph16b0 Solution: Assume that two deuteron nuclei in the core of fusion reactor…

To determine the minimum temperature required for two deuteron nuclei to reach a separation of $4 \times 10^{-15}$ m, we need to equate the initial kinetic energy of the deuterons with the potential energy at the specified separation distance.

The initial kinetic energy of each deuteron is given as :

$KE = 1.5 \, k \, T$

Since there are two deuterons, the total kinetic energy is :

$KE_{\text{total}} = 2 \times 1.5 \, k \, T = 3 \, k \, T$

The Coulomb potential energy (PE) between two deuterons at a separation distance $r = 4 \times 10^{-15}$ m is given by :

$PE = \frac{e^2}{4 \pi \epsilon_0 r}$

Using the given value:

$\frac{e^2}{4 \pi \epsilon_0} = 1.44 \times 10^9 \, \text{eV} \cdot \text{m}$

So, the potential energy becomes :

$PE = \frac{1.44 \times 10^9 \, \text{eV} \cdot \text{m}}{4 \times 10^{-15} \, \text{m}}$

Simplifying :

$PE = \frac{1.44 \times 10^9}{4 \times 10^{-15}} \, \text{eV}$

$PE = 0.36 \times 10^{24} \, \text{eV}$

Now, we equate the total kinetic energy to the potential energy to find the temperature $T$ :

$3 \, k \, T = 0.36 \times 10^{24} \, \text{eV}$

Solving for $T$ :

$T = \frac{0.36 \times 10^{24}}{3 \times 8.6 \times 10^{-5}} \, \text{K}$

$T = \frac{0.36 \times 10^{24}}{2.58 \times 10^{-4}} \, \text{K}$

$T = 1.395 \times 10^{9} \, \text{K}$

This temperature is in the range :

Option A: 1.0 \times {10^9}K < T < 2.0 < {10^9}K

So, the correct answer is :

Option A : 1.0 \times {10^9}K < T < 2.0 < {10^9}K