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AP EAMCET · PHYSICS · Electrostatics

The maximum potential energy due to electrostatic repulsion between two hydrogen nucleus is nearly (radius of the nucleus =1.1 Fermi) 14πε0=9×109 N m2 C-2

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

Answer & Solution

Correct Answer

(A) 0.65MeV

Step-by-step Solution

Detailed explanation

The separation between two hydrogen nucleus should be equal to the diameter of nucleus, ⇒r=2r0=2×1.1×10-15 m so potential energy of the system, U=14πε0q1q2r=9×109×1.6×10-1922×1.1×10-15=0.65 MeV
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