Q. 183.9( 13 Votes )

# 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?

Answer :

Given,

The distance between electron-proton of hydrogen atom = 0.53 Å = 0.53 × 10^{-10}m

Charge on electron, q_{1} = - 1.6 × 10^{-19} C

Charge on proton, q_{2} = 1.6 × 10^{-19} C

(a) Potential energy at infinity is Zero.

Potential energy of a system,

= Potential energy at infinity – Potential energy at distance d

= 0 -

Where, ϵ_{0} = permittivity of space

and, =

∴ Potential energy,

⇒ V = -43.7 × 10^{-19} J

On converting into electron volts we get,

⇒

⇒ V = -27.2 eV

(∵ 1eV = 1.6 × 10^{-19} J)

(b) Given, Kinetic energy is half of the potential energy in (a).

∴ Kinetic energy, K = 0.5 × (27.2) = 13.6 eV

Thus, Total energy of electron = -27.2 + 13.6 = -13.6 eV

Amount of work require to free the electron, A = Increase in energy the electron.

∴ A = 0-(-13.6) = 13.6 eV

(c) If we take potential energy Zero at 1.06 Å separation, then the potential energy of the system,

P =

⇒P = -21.74 × 10^{-19} J

In electron volts, the potential energy is given as,

P = -13.585 eV.

(1eV = 1.6 × 10^{-19} J)

∴ The amount of work done to free the electron in this case,

W = -27.17- (-13.585) = -13.585 eV

Rate this question :

A charge 'q' is moved from a point A above a dipole of dipole moment 'p' to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.

Physics - Board Papers