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Question: If two deuterium nuclei get close enough together, the attraction of the strong nuclear force will f...

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{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. 1.2×10−6−ms−11.2 \times {10^{ - 6}} - m{s^{ - 1}}
B. 1.2×107ms−11.2 \times {10^7}m{s^{ - 1}}
C. 3.8×106ms−13.8 \times {10^6}m{s^{ - 1}}
D. 8.72×106ms−18.72 \times {10^6}m{s^{ - 1}}

Explanation

Solution

Hint We know about columbic force. We will write the kinetic energy equation because of its velocity. And we will write the energy due to their attraction. After comparing both the equations we will get the velocity of the two nuclei.

Complete step by step solution
Let the value of two charges be q1 {q_{1\,}} and q2{q_2}
The distance between them is r that is the range of nuclear force is 10−15{10^{ - 15}} m.
So, they got close enough to each other so the attraction will be due to coulombic force.
Potential energy interaction between two charges separated by the distance is mathematically written as follows:
KE=q1q24πζ0rKE = \dfrac{{{q_1}{q_2}}}{{4\pi {\zeta _0}r}}
The kinetic energy will be,
=12mv2= \dfrac{1}{2}m{v^2}
Here m is the mass of the charge.
Let the velocity of nuclei be v coming towards each other for head collision to fuse.
Equating the energy we get,
mv22=q1q24πζ0r\dfrac{{m{v^2}}}{2} = \dfrac{{{q_1}{q_2}}}{{4\pi {\zeta _0}r}}
v2=1.6×10−19×1.6×10−19×9×109×210−15×2×1.67×10−27{v^2} = \dfrac{{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=1.2×107ms−1v = 1.2 \times {10^7}m{s^{ - 1}}
Hence the required solution is v=1.2×107 ms−1v = 1.2 \times {10^7}\,m{s^{ - 1}} (Option- “B”)

Note
So, one may think about how to calculate the velocity of the two nuclei. There exists also a coulombic force. So due to the two charges, potential energy exists. Due to their motion we can easily get their kinetic energy. So, this is a very easy problem. We just have to equate the potential energy with kinetic. Lastly taking square root both sides we get the solution is 1.2×107 ms−11.2 \times {10^7}\,m{s^{ - 1}} (Option- “B”)