The radius of a nucleus is given by r_0 A^{1 | vedclass

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Question

(a)

Taking protons at dimetrically opposite points of nucleus,

Force of electrostatic repulsion,

$F=\frac{k(e)(e)}{r^2}=\frac{k e^2}{\left(r_

Vedclass · Accepted Answer

$10^2$

Vedclass · Answer

(a)

Taking protons at dimetrically opposite points of nucleus,

Force of electrostatic repulsion,

$F=\frac{k(e)(e)}{r^2}=\frac{k e^2}{\left(r_0 A^{1 / 3}ight)^2}-\frac{k \cdot e^2}{r_0^2 \cdot A^{2 / 3}}$

Substituting values in above equation, we get

$F=\frac{9 	imes 10^9 	imes\left(16 	imes 10^{-19}ight)^2}{\left(13 	imes 10^{-15}ight)^2(206)^{23}}$

$=0.039 	imes 10^2 \,N$

The radius of a nucleus is given by $r_0 A^{1 / 3}$, where $r_0=1.3 \times 10^{-15} \,m$ and $A$ is the mass number of the nucleus. The lead nucleus has $A=206$. The electrostatic force between two protons in this nucleus is approximately ................ $N$

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