In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

The distance between electron-proton of a hydrogen atom, d = 0.53 A

Charge on an electron, q₁ = -1.6 x 10^-19 C Charge on a proton, q₂ = +1.6 x 10^-19 C

Ans (a).

Potential at infinity is zero.

Potential energy of the system, p-e

= Potential energy at infinity – Potential energy at distance, d

= 0 – (q₁-q₂)/4πε₀ d

Where, ε₀ is the permittivity of free space and

1/4πε₀ =9x 10⁹ Nm²C⁻²

Therefore Potential Energy = 0- (9x 10⁹) x (1.6 x 10⁻¹⁹)² /(0.53 x 10¹⁰)

= -43.7x 10⁻¹⁹ J , (Since 1.6 x 10⁻¹⁹ J = 1 eV)

Therefore  Potential energy =-43.7×10⁻¹⁹ = -(43.7 x 10⁻¹⁹)/(1.6 x 10⁻¹⁹) =-27.2 eV

Therefore, the potential energy of the system is -27.2 eV.

Ans (b).

Kinetic energy is half of the magnitude of potential energy.

Kinetic energy = 1/2 x(-27.2) = 13.6 eV

Total energy = 13.6 – 27.2 = 13.6 eV

Therefore, the minimum work required to free the electron is 13.6 eV.

Ans (c).

When zero of potential energy is taken, d = 1.06 A

Therefore Potential energy of the system = Potential energy at d₁- Potential energy at d

₌ (q₁q₂)/4πε₀ d₁   –   27.2 eV

= (9 x 10⁹) x (1.6 x 10⁻¹⁹)²  / ( 1.06 x 10⁻¹⁰)   -27.2 eV

=21.73 x 10⁻¹⁹ J – 27.2 eV =

=13.58 eV – 27.2 eV= – 13.6 eV