AP EAMCET201825 Apr 2018Morning ShiftPhysicsElectrostaticsActual
The maximum potential energy due to electrostatic repulsion between two hydrogen nucleus is nearly (radius of the nucleus = 1 . 1 Fermi) 1 4 π ε 0 = 9 × 10 9 N m 2 C - 2
Options
- A0 . 65 MeV
- B2 . 09 MeV
- C3 . 31 MeV
- D0 . 92 MeV
Correct answer
A. 0 . 65 MeV
Step-by-step solution
The separation between two hydrogen nucleus should be equal to the diameter of nucleus, ⇒ r = 2 r 0 = 2 × 1 . 1 × 10 - 15   m so potential energy of the system, U = 1 4 π ε 0 q 1 q 2 r = 9 × 10 9 × 1 . 6 × 10 - 19 2 2 × 1 . 1 × 10 - 15 = 0 . 65   MeV