Ncert Exercises & Solutions

Question 2.18: 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 an infinite separation of the electron from the proton.

(b) What is the minimum work required to free the electron, given that it's kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

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

Charge on an electron,

q_{1} = - 1.6 x 10^{- 19} C

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

(a) Potential at infinity is zero.

The potential energy of the system, p - e = Potential energy at infinity - Potential energy at distance, d

Where, is the permittivity Of free space -9xlO" Nm2 9x109 = —43.7 x IO'IOJ :. Potential energy = O— Since ev, —43.7 x | ()-10 . Potential energy = —43.7 x 10•10 1.6* ION = -272 ev

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

(b) Kinetic energy is half Of the magnitude Of potential energy.

Kinetic energy = \frac{1}{2} \times (- 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,

∴

The potential energy of the system = Potential energy at d₁ - Potential energy at d

\begin{array}{l} = \frac{q_{1} q_{2}}{4 {πϵ}_{0} d_{1}} - 27 .2 eV \\ = \frac{9 \times 10^{9} \times {(1 .6 \times 10^{- 19})}^{2}}{1 .06 \times 10^{- 10}} - 27 .2 eV \\ = 21 .73 \times 10^{- 19} J - 27 .2 eV \\ = 13 .58 eV - 27 .2 eV \\ = - 13 .6 eV \end{array}

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