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Karnataka Board PUCPUC Science 2nd PUC Class 12
पाठ्यपुस्तक उत्तर१६९६४
MCQ ऑनलाइन अभ्यास परीक्षा२२
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पाठ्यक्रम

Electron Orbits

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Topics

Estimated time: 7 minutes
CBSE: Class 12

Introduction

The Rutherford nuclear model of the atom pictures the atom as an electrically neutral sphere consisting of a very small, massive, and positively charged nucleus at the centre, surrounded by revolving electrons in their respective dynamically stable orbits. The electrostatic force of attraction between the revolving electron and the nucleus provides the centripetal force required to keep the electron in orbit.

CBSE: Class 12

Force Relation

For a dynamically stable orbit in a hydrogen atom,

Fe = Fc
\[\frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}=\frac{mv^2}{r}\]

Thus, the relation between the orbital radius and the electron velocity is:

r = \[\frac{e^2}{4\pi\epsilon_0mv^2}\]​

These relations are given directly in the provided content.

CBSE: Class 12

Energy of the Electron

The kinetic energy (K) and electrostatic potential energy (U) of the electron in a hydrogen atom are given as:

K = \[\frac{1}{2}mv^2=\frac{e^2}{8\pi\epsilon_0r}\]
U = -\[\frac{e^2}{4\pi\epsilon_0r}\]

The negative sign in (U) signifies that the electrostatic force is in the negative (r) direction.

CBSE: Class 12

Total Energy

The total energy (E) of the electron in a hydrogen atom is:

E = K + U
E = \[\frac{e^2}{8\pi\epsilon_0r}-\frac{e^2}{4\pi\epsilon_0r}\]
E = −\[\frac{e^2}{8\pi\epsilon_0r}\]

The total energy of the electron is negative. This implies that the electron is bound to the nucleus. If (E) were positive, the electron would not follow a closed orbit around the nucleus.

CBSE: Class 12

Example

It has been experimentally found that 13.6 eV of energy is required to separate a hydrogen atom into a proton and an electron. The example asks to compute the electron's orbital radius and velocity in a hydrogen atom.

Given

  • The total energy of an electron in a hydrogen atom is -13.6 eV.

  • \(-13.6 , \text{eV} = -13.6 \times 1.6 \times 10^{-19} , \text{J} = -2.2 \times 10^{-18} , \text{J}\).

Using Eq.

E =\[-\frac{e^2}{8\pi\epsilon_0r}=-2.2\times10^{-18}\mathrm{J}\]

This gives the orbital radius:

r = −\[\frac{e^2}{8\pi\epsilon_0E}\]
r = 5.3  ×10−11 m

Velocity of the electron

The velocity of the revolving electron can be computed with (m = 9.1 × 10-31, kg).

v = \[\frac{e}{\sqrt{4\pi\epsilon_0mr}}\]​
v = 2.2 × 106 m/s
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