Question

Two $C{{u}^{64}}$ nuclei touch each other. The electrostatics repulsive energy of the system will be
(A)- 0.788 Mev
(B)- 7.88 Mev
(C)- 126.15 MeV
(D)- 788 MeV

Answer 1

Hint : Here we are given Copper nuclei and we know that the radius of nucleus is given by the formula $R={{R}_{0}}{{(A)}^{1/3}}$ , where R is the radius of the nucleus and ${{R}_{0}}$ is a constant and A is the atomic mass of the nucleus. After finding the radius, since the two nuclei touch each other, the distance from the center of one to the other will be twice of the radius. So, after finding the distance between them we can easily find the potential energy.

Complete Step By Step Answer:
Using $R={{R}_{0}}{{(A)}^{1/3}}$ , here ${{R}_{0}}=1.2fm$ and A for copper is given as 64. So, putting the value in the equation we get,
$R=1.2{{(64)}^{1/3}}=1.2\times 4=4.8fm=4.8\times {{10}^{-15}}m$
The potential energy is given by the formula: 2RU
$=\dfrac{k{{q}^{2}}}{r}$
$=\dfrac{9\times {{10}^{9}}\times (1.6\times {{10}^{-19}}\times 29)}{2\times 4.8\times {{10}^{-15}}\times 1.6\times {{10}^{-19}}}$
$=126.15Mev$
So, the potential energy comes out to be 126.15 Mev.
Hence, the correct option is (C).

Note :
While finding out the potential energy we have multiplied the value of charge by 29 because there are 29 protons inside the nucleus. 29 is also the atomic number of copper. The atomic number for a neutral atom is equal to the number of protons and electrons in that atom.
Also, the answer has been expressed in units of MeV where $1Mev={{10}^{6}}ev$
Also, to convert Joules into eV we divide by $1.6\times {{10}^{-19}}$