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AP EAMCET · PHYSICS · Electrostatics

The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus
\[
=1.1 \text { fermi })\left[\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right]
\]

  1. A 0.65 MeV
  2. B 2.09 MeV
  3. C 3.31 MeV
  4. D 0.92 MeV
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Answer & Solution

Correct Answer

(A) 0.65 MeV

Step-by-step Solution

Detailed explanation

Potential energy due to two charges \(=\frac{k q_1 q_2}{d}\). For hydrogen atom, \(q_1=q_2=1.6 \times 10^{-19} \mathrm{C}\)…
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