AP EAMCET201823 Apr 2018Morning ShiftPhysicsElectrostaticsActual
The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus =1.1 fermi ) [ 1 4 ₀ =9 10^9 Nm ^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
Potential energy due to two charges = k q₁ q₂ d . For hydrogen atom, q₁=q₂=1.6 10⁻¹⁹ C aligned U & = 9 10^9 (1.6 10⁻¹⁹ )^2 2.2 10⁻¹⁵ = & 10.47 10⁻¹⁴ ~J = 10.47 10⁻⁴ 1.6 10⁻¹⁹ eV & =6.5 10^5 eV =0.65 MeV aligned