Q172 / 200 · Manipal (MET) 2014 · Physics

Two $\mathrm{Cu}^{64}$ nuclei touch each other. The electrostatics repulsive energy of the system will be

  • A
    0.788 MeV
  • B
    7.88 MeV
  • Correct answer
    C
    126.15 MeV
  • D
    788 MeV

Solution

Radius of each nucleus $R=R_0(A)^{1 / 3}=1.2(64)^{1 / 3}$ $=4.8 \mathrm{~m}$ Distance between two nuclei $(r)=2 R$ So, potential energy $U=\frac{k \cdot q^2}{r}$ $=\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19} \times 29\right)^2}{2 \times 4.8 \times 10^{-15} \times 1.6 \times 10^{-19}}=126.15 \mathrm{MeV}$

Hence, the correct answer is option: 'C'.

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