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 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 separations?

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alok kumar singh | Contributor-Level 10

3 months ago

2.18 The distance between electron – proton of a hydrogen atom, d = 0.53 Å = 0.53 $\times 10^{- 10}$ m

Charge of an electron, $q_{1}$ = $-$ 1.6 $\times 10^{- 19}$ C

Charge of a proton, $q_{2}$ = 1.6 $\times 10^{- 19}$ C

Potential at infinity = 0

Potential energy of the system = Potential energy at infinity – Potential energy at a distance d

= 0 - $\frac{q_{1} q_{2}}{4 π ε_{0} d}$ , where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ == 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

= 0 - $\frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{4 π \times 8.854 \times 10^{- 12} \times 0.53 \times 10^{- 10}}$ = - 4.34 $\times 10^{- 18}$ J = $\frac{4.34 \times 10^{- 18}}{1.6 \times 10^{- 19}}$ = - 27.13 eV

(Since 1 eV = 1.6 $\times 10^{- 19}$ J)

Kinetic energy = $\frac{1}{2}$ of potential energy = $\frac{1}{2}$ $\times -$ 27.13 eV = -13.57 eV

Total energy = - 13.57 – (-27.13) = 13.57 eV

Ther

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2.18 The distance between electron – proton of a hydrogen atom, d = 0.53 Å = 0.53 $\times 10^{- 10}$ m

Charge of an electron, $q_{1}$ = $-$ 1.6 $\times 10^{- 19}$ C

Charge of a proton, $q_{2}$ = 1.6 $\times 10^{- 19}$ C

Potential at infinity = 0

Potential energy of the system = Potential energy at infinity – Potential energy at a distance d

= 0 - $\frac{q_{1} q_{2}}{4 π ε_{0} d}$ , where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ == 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

(Since 1 eV = 1.6 $\times 10^{- 19}$ J)

Kinetic energy = $\frac{1}{2}$ of potential energy = $\frac{1}{2}$ $\times -$ 27.13 eV = -13.57 eV

Total energy = - 13.57 – (-27.13) = 13.57 eV

Therefore, the minimum work required to free the electron is 13.57 eV

In case of zero potential energy, $d_{1}$ = 1.06 Å = 1.06 $\times 10^{- 10}$ m

Potential energy of the system = Potential energy at $d_{1}$ - Potential energy at d

= $\frac{q_{1} q_{2}}{4 π ε_{0} d}$ $\times \frac{1}{1.6 \times 10^{- 19}}$ - 27.13 eV = $\frac{{(1.6 \times 10^{- 19})}^{2}}{4 π \times 8.854 \times 10^{- 12} \times 1.06 \times 10^{- 10}} \times \frac{1}{1.6 \times 10^{- 19}} -$ 27.13 = 13.56 -27.13 eV= -13.56 eV

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