Question

The radius of a hypothetical nucleus (atomic number$=79$) is about $7\times {10}^{-15}$ m. Assuming that charge distribution is uniform, the electric field at the surface of the nucleus is :

• A. $2.9\times {10}^{-21}$
• B. $2.32\times {10}^{21}$
• C. $4.64\times {10}^{21}$
• D. $0$

Answer 1

The correct option is B $2.32\times {10}^{21}$

The expression of electric field at surface of sphere of radius a is given by, $E=\frac{kq}{a^2}$

Here, charge $q=79e=79\times (1.6\times {10}^{-19})$ and $a=7\times {10}^{-15}$

So, $E=\frac{(9\times {10}^9)\times (79\times 1.6\times {10}^{-19})}{(7\times {10}^{-15}{)}^2}=2.32\times {10}^{21}N/C$