Atomic Physics- Rutherford Atom Model

Rutherford Atom Model

The essential features of Rutherford’s nuclear model of the atom or planetary model of the atom are as follows:

1.Every atom consists of a tiny central core, called the atomic nucleus, in which the entire positive charge and almost entire mass of the atom are concentrated.

2.The size of nucleus is of nucleus is of the order of10^-15m, which is very small as compared to the size of the atom which is of the order of10^-10m.

3.The atomic nucleus is surrounded by certain number of electrons. As atom on the whole is electrically neutral, the total negative charge of electrons surrounding the nucleus is equal to total positive charge on the nucleus.

4.These electrons revolve around the nucleus in various circular orbits as do the planets around the sun. The centripetal force required by electron for revolution is provided by the electrostatic force of attraction between the electrons and the nucleus.

Electron orbits

IfF_c=centripetal force required to keep a revolving electron in orbit.

F_e= electrostatic force of attraction between the revolving electron and the nucleus then

For a dynamically stable orbit in a hydrogen atom,

F_c= F_e

m (v²/r) = (e) (e)/4∏?₀r²(1)

r = e²/4∏?₀mv²(2)

K.E.of electron in the orbit,K = ½ mv²

Using (2),K = e²/8∏?₀r

电子的势能orbit

U = (e) (-e)/ 4∏?₀r = -e²/4∏?₀r

Negative sign indicates that revolving electron is bound to the positive nucleus.

Therefore, total energy of electron in orbit of hydrogen atom is negative. Hence, the electron is bound to the nucleus.

Example:calculate the impact parameter of a MeV alpha particle scattered by10°when it approaches a gold nucleus. TakeZ = 79for gold.

Solution:here,K.E. = ½ mv²= 5 MeV

= 5 × 1.6 ×10^-13J,

θ = 10°, Z = 79, b=?

Asb = Ze²(cot θ/2)/4∏?₀(1/2 mv²)

b = 9 × 10⁹× 79 (1.6 × 10^-19)²cot 5°/5 × 1.6 × 10^-13

= 9 × 79 × 1.6 × 1.6 × 10^-16/8 × 0.0875 (cot 5° = 0.0875)

b = 2.6 × 10^-13m

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