Question

Suppose two deutrons must get as close as ${10}^{-14}m$ in order for the nuclear force to overcome the repulsive electrostatic force. The height of the electrostatic barrier is nearest to

• A. $0.14MeV$
• B. $2.3MeV$
• C. $1.8\times 10MeV$
• D. $0.56MeV$

Answer 1

The correct option is A $0.14MeV$

Electrostatic barrier developed is given by $\frac{k{q}^2}{r}$, where value of $k$ is $8.98\times {10}^{9}N{m}^2.C^{-2}$
Dueterium is one proton + one neutron
Therefore, charge of duetrium is $1.602\times {10}^{-19}C$.
Hence, barrier developed is $=\frac{8.98\times {10}^{9}N{m}^2.C^{-2}\times {(1.602\times {10}^{-19}C)}^2}{{10}^{-14}m}$$=2.3\times {10}^{-14}N=0.14MeV$