AP EAMCET · PHYSICS · Electrostatics
The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus
\[
=1.1 \text { fermi })\left[\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right]
\]
- A 0.65 MeV
- B 2.09 MeV
- C 3.31 MeV
- D 0.92 MeV
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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