**CBSE NOTES CLASS 12 PHYSICS **

**CHAPTER 12 ATOMS**

**Rutherford’s atomic model**

On the basis of α–particle scattering experiment, Rutherford made following observations.

The entire positive charge and almost entire mass of the atom is concentrated at its centre in a very tiny region of the order of 10^{-15} m, called nucleus.

He proposed the following model for the structure of atom.

- The negatively charged electrons revolve around the nucleus in different orbits.
- The total positive charge on the nucleus is equal to the total negative charge on electron. Therefore atom as a whole is neutral.
- The electrons revolve around the nucleus. The centripetal force required by electron for revolution is provided by the electrostatic force of attraction between the electrons and the nucleus.

**Electrostatic force between the nucleus and α–particle**

When the α–particle approaches a nucleus with atomic number Z (+ve charge Ze) the magicnitude of this **electrostatic force**

$$\mathrm{F}=\frac{1}{4\mathrm{\pi}{\mathrm{\varepsilon}}_{\mathrm{o}}}\frac{\left(2\mathrm{e}\right)\left(\mathrm{Z}\mathrm{e}\right)}{{\mathrm{r}}^{2}}$$

**Distance of closest approach**

$${\mathrm{r}}_{\mathrm{o}}\mathrm{}\mathrm{}=\mathrm{}\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\mathrm{}\frac{{2\mathrm{Z}\mathrm{e}}^{2}}{{\mathrm{E}}_{\mathrm{K}}}$$

where, E_{K} = kinetic energy of the α-particle.

**Impact parameter**

The impact parameter is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus.

$$\mathrm{b}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\mathrm{}\frac{{\mathrm{Z}\mathrm{e}}^{2}\mathrm{cot}\frac{\theta}{2}}{{\mathrm{E}}_{\mathrm{K}}}\mathrm{}$$

**Rutherford’s scattering formula**

$$\mathrm{N(\theta )}=\mathrm{}\frac{{\mathrm{N}}_{\mathrm{i}}\mathrm{n}\mathrm{t}\mathrm{}{\mathrm{Z}}^{2}{\mathrm{e}}^{4}}{{\left(8\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}\right)}^{2}\mathrm{}{\mathrm{r}}^{2}{\mathrm{E}}^{2}\mathrm{}{\mathrm{s}\mathrm{i}\mathrm{n}}^{4}\frac{\theta}{2}}$$

where,

N(θ) = number of α–particles with scattering angle θ,

N_{i} = total number of α-particles reaching the screen,

n = number of atoms per unit volume in the foil,

Z = atomic number of the foil,

E = kinetic energy of the alpha particles and

t = foil thickness

**Limitations of Rutherford’s atomic model**

**About the stability of atom -**According to Maxwell’s electromagnetic wave theory electron should emit energy in the form of electromagnetic wave during its orbital motion. Therefore radius of orbit of electron will decrease gradually and ultimately it will fall in the nucleus.

**About the line spectrum -**Rutherford atomic model cannot explain atomic line spectrum.