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Question: Suppose two deuterons must get as close as \[{10^{ - 14}}m\] in order for the nuclear force to overc...

Suppose two deuterons must get as close as 10−14m{10^{ - 14}}m in order for the nuclear force to overcome the repulsive electrostatic force. The height of the electrostatic barrier is nearest to
(A)0.14MeV(A)0.14MeV
(B)2.3MeV(B)2.3MeV
(C)1.8×10MeV(C)1.8 \times 10MeV
(D)0.56MeV(D)0.56MeV

Explanation

Solution

The height of the potential barrier means the electrostatic potential energy between the two deuterons. So, we use the equation of electrostatic potential energy between two charged particles to find the height of the electrostatic barrier.

Formula used:
U=kq2rU = \dfrac{{k{q^2}}}{r}
Where UU= potential energy or barrier, qq is the charge of deuterium and rr= distance between the two charges.

Complete step-by-step solution:
Electrostatic barrier developed is given by U=kq2rU = \dfrac{{k{q^2}}}{r}
Where the value of k=9×109Nm2.C−2k = 9 \times {10^9}N{m^2}.{C^{ - 2}}
Since the deuterons are the charged particles whose magnitude of the charge is equal to that of an electron.
Therefore, q=1.6×10−19Cq = 1.6 \times {10^{ - 19}}C
We know that, suppose two deuterons must get as close as 10−14m{10^{ - 14}}m in order for the nuclear force to overcome the repulsive electrostatic force.
Therefore r=10−14mr = {10^{ - 14}}m
Substituting all the values in the equation U=kq2rU = \dfrac{{k{q^2}}}{r} and simplify
U=(9×109)×(1.6×10−19)210−14U = \dfrac{{\left( {9 \times {{10}^9}} \right) \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{{{10}^{ - 14}}}}
⇒U=2.3×10−14N=0.14MeV\Rightarrow U = 2.3 \times {10^{ - 14}}N = 0.14MeV
Hence, option A is correct.
Note: Always remember that there is a difference between the electrostatic potential energy and electrostatic potential. The amount of work done in moving a unit positive charge between infinity and that point without any acceleration against the electric force is called Electrostatic potential at any point in an electric field. Every point around a source charge is characterized with electric potential is given by-
V=kqrV = \dfrac{{kq}}{r}
Where V= Electric potential
If instead of bringing a unit positive, we bring a charge 'qq' from infinity to that point, work done WW in doing so is given by-
W=q×VW = q \times V
⇒W=kq2r\Rightarrow W = \dfrac{{k{q^2}}}{r}
This work is termed as 'electrostatic potential energy of the charge qq in the field of charge ‘qq'. We can also say that the electric potential at any point in an electric field is the potential energy of a unit positive charge placed at that point.