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12.Atoms
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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 ................ $N$
A
$10^2$
B
$10^7$
C
$10^{12}$
D
$10^{17}$
(KVPY-2016)
Solution
(a)
Taking protons at dimetrically opposite points of nucleus,
Force of electrostatic repulsion,
$F=\frac{k(e)(e)}{r^2}=\frac{k e^2}{\left(r_0 A^{1 / 3}\right)^2}-\frac{k \cdot e^2}{r_0^2 \cdot A^{2 / 3}}$
Substituting values in above equation, we get
$F=\frac{9 \times 10^9 \times\left(16 \times 10^{-19}\right)^2}{\left(13 \times 10^{-15}\right)^2(206)^{23}}$
$=0.039 \times 10^2 \,N$
Standard 12
Physics
