KVPY SA Exam 2016 Question Paper

Question : 25 of 80

Marks: +1, -0

The radius of a nucleus is given by

r_0 A^{1/3}

where

r_0 = 1.3 × 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

Solution:

r=r_0\A^{1\/3}

r_0=1.3×10^{-15}

F=1/{4π∈_0}×{q_1q_2}/r^2

F={9×10^9×1.6×10^{-19}×1.6×10^{-19}}/{r_0^2\A^{2\/3}}

F={9×10^9×1.6×10^{-19}×1.6×10^{-19}}/{1.3^2×10^{-30}×(206)^{2\/3}}

={23.04×10^39×10^{-38}}/{1.69×34.81}

= 3.91 Newton

= 0.039 × 10^2

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