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`        Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, 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 × 10-15 m is in the range`
7 years ago

28 Points
```										Dear Ankita Shakya,
Ans:- The initial energy is totally linetic and the total energy is=2(1.5kT)=3kT and initially there is no coulombic P.E.
Then finally the total energy must be converted to the potential energy as the temp is minimum( assuming the final temp to be zero). Hece from energy equation we get
Kq²/r=initial KE
Solving we get,
T=(5.12×10^-14)/k(ans)

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Ankita Shakya !!!

Regards,
```
7 years ago
28 Points
```										Dear STUDENT NAME,
In the previous solution the answer was wrong due to some calculation mistake please note the correct answer. The correct Temperature is T=(1.92×10^-14)/k Kelvin where k is a constant
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best STUDENT NAME !!!

Regards,
```
7 years ago
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