The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus =1.1 fermi) [(1/4 π ε0)=9 × 109 Nm 2 C -2]

Question

The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus =1.1 fermi) [(1/4 π ε0)=9 × 109 Nm 2 C -2] (A) 0.65 MeV (B) 2.09 MeV (C) 3.31 MeV (D) 0.92 MeV

Tardigrade Admin · Accepted Answer

Potential energy due to two charges =(k q1 q2/d). For hydrogen atom, q1=q2=1.6 × 10-19 C U =(9 × 109 ×(1.6 × 10-19)2/2.2 × 10-15) [∴ d=2s] = 10.47 × 10-14 J =(10.47 × 10-4/1.6 × 10-19) eV =6.5 × 105 eV =0.65 MeV

Q. The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus =1.1 fermi) [4πε0​1​=9×109Nm2C−2]

0.65 MeV

2.09 MeV

3.31 MeV

0.92 MeV

Potential energy due to two charges =dkq1​q2​​. For hydrogen atom, q1​=q2​=1.6×10−19C U=2.2×10−159×109×(1.6×10−19)2​[∴d=2s] =10.47×10−14J=1.6×10−1910.47×10−4​eV =6.5×105eV=0.65MeV

Q. The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly
(radius of the nucleus $= 1.1$ fermi) $[\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}]$