Ncert Exercises & Solutions

In a hydrogen atom with the electron and proton bound at a distance of \(0.52~\mathring{\text {A}},\) estimate the potential energy of the system in \(\text{eV}\) (with zero at infinite separation) and the minimum work required to free the electron (given its kinetic energy is half the magnitude of this potential energy), and then determine the potential energy if the zero of potential energy is taken at a separation of \(1.06~\mathring{\text A}.\)
1. \((-13.6~\text {eV},~6.8~\text {eV})~\text{and}~(-6.8~\text{eV})\)
2. \((-27.2~\text {eV},~13.6~\text {eV})~\text{and}~(-13.6~\text{eV})\)
3. \((27.2~\text {eV},~-13.6~\text {eV})~\text{and}~(13.6~\text{eV})\)
4. \((13.6~\text {eV}, -6.8~\text {eV})~\text{and}~(6.8~\text{eV})\)

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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