Question

Two ${}_{29}C{u}^{64}$ nuclei almost touch each other. The electronics repulsive energy of the system will be $(R_0=1\text{fm})$

• A. $0.788\text{MeV}$
• B. $7.88\text{MeV}$
• C. $126.15\text{MeV}$
• D. $788\text{MeV}$

Answer 1

The correct option is C $126.15\text{MeV}$

Radius of each nucleus,

$R=R_0(A{)}^{1/3}=1.2(64{)}^{1/3}=4.8\text{fm}=4.8\times {10}^{-15}\text{m}$

Distance between two nuclei $\left(r\right)=2R=2\times 4.8\times {10}^{-15}\text{m}$

So, potential energy

$U=\frac{k{q}^2}{r}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}\times 29{)}^2}{2\times 4.8\times {10}^{-15}}$

Potential energy in $\left(\text{eV}\right)$,

$U=\frac{k{q}^2}{r{q}_{electron}}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}\times 29{)}^2}{2\times 4.8\times {10}^{-15}\times 1.6\times {10}^{-19}}$

$=126.15\times {10}^6\text{eV}=126.15\text{MeV}$

Hence, $\left(\text{C}\right)$ is the correct answer.