The Nuclear Atom, Rutherford Scattering and it's Conclusion.

The Nuclear Atom:

Before we can make further progress in relating the energy levels of an atom to its internal structure, we need to have a better idea of what the inside of an atom is like. We know that atoms are much smaller than the wavelengths of visible light, so there is no hope of actually seeing an atom using that light. But we can still describe how the mass and electric charge are distributed throughout the volume of the atom.

Here's where things stood in 1910. J.J. Thomson had discovered the electron and measured its charge-to-mass ratio (e/m) in 1897; by 1909, Millikan had completed his first measurements of the electron charge -e. These and other experiments showed that almost all the mass of an atom had to be associated with the positive charge, not with the electrons. It was also known that the overall size of atom is of the order of \(10^{-10}\) m and that all atoms except hydrogen contain more than one electron. What was not known was how the mass and charge were distributed within the atom. Thomson had proposed a model in which the atom consisted of a sphere of positive charge, of the order of \(10^{-10}\) m in diameter, with the electrons embedded in it like raisins in a more or less spherical muffin.

Thomson's Atomic Model:

According to Thomson atomic model, "The atom might be regarded as a sphere of positive charge in which negatively charged electron were embeded in it" as shown in figure 9.1.

He conceived that his attomic model as a sphere of size \(10^{-10}\) m.Ah atom is electrically neutral, just a sufficient number of negatively charged electron was supposed to be embedded with in the atom to balance the positive charge. This atomic model is also called as plum pudding model.

Limitations of Thomson's atomic model

1. According to tis model, Hydrogen can give rise to a single spectral line while experimentally it was observed that the Hydrogen spectrum contained several series with several lines in each series. Thus this mode is defective.

2. Rutherford's alpha particle scattering experiments showed that the entire positive charge in atom is concentrated to very small region at centre which could not explained by Thomson's atomic model.

Rutherford's Scattering and it's conclusion:

Rutherford's \(\alpha\)-particle scattering experiment

\(\alpha\)-particles are doubly charged helium atoms that have lost both the electrons. Thus, the mass of \(\alpha\)-particle is four times the mass of a hydrogen atom and a positive charge two times the charge of proton.

A collimated beam of \(\alpha\)-particles coming from the radioactive source redium were made incident on the thin gold foil G (about 1 \(\mu\) thick ) beyound what a movable zinc sulphide screen was place. The \(\alpha\)-particles scattered from the gold foil at angles varying from 0 to \(180^\circ\) when foil on the zns screen produces, flashes of light which are detected by microscope M, which is shown in above figure 9.2.

Result of Rutherford\(\alpha\)-particle:

1. Most of alpha-particles passed along straight line through the gold foil indicating thereby that the atoms are hollow.

2. Some of \(\alpha\)-particle were scattered through small angles.

3. A few of them ( one of about 8000 particles) were scattered by angle greater than \(90^\circ\).

4. Sometimes \(\alpha\)-particle went back along the direction from which it came at an angle of scattering of \(180^\circ\).

Theory of Rutherford \(\alpha\)-particle:

Assumption of Rutherford \(\alpha\)-particle :-

1. The entire positive charges and the most of the mass of atoms are concentrated in a very small and highly dense region called nucleus. The size of the atom is of the order of \(10^{-10}\) and that of the nucleus is \(10^{-55}\)m.

2. To explain the stability of the atom against the falling of the electrons into the nucleus under the electrostatic attraction he postulated that like solar system, electron revolve around the nucleus in circular orbits, so that centrifugal force due to the rotation of the electron is balanced by the electrostatic force of attraction between the nucleus and the electrons.

Theory of \(\alpha\) - Particle Scattering

Let us consider mass of \(\alpha\) -particle is m and charge is 2e approaching toward a target nucleus of charge ze, is scattered. The path followed by the \(\alpha\) - particle is hyperbolic. The scattering is due to the coloumb electrostatic forces between \(\alpha\) -particle and the positive charge of the nucleus. This force is varies as \(\frac{1}{r^2}\). Where , r is the instantaneous seperation between \(\alpha\) -particle and the target nucleus.

Some parameters:

1. Angle of scattering (\(\theta\)): The angle between the asymptotic direction of approach of \(\alpha\) -particle and the asymptotic direction in which it recedes is called angle of scattering. Here \(theta\) is the angle of scattering.

2. Impact parameter (b) : The minimum distance to which \(\alpha\) -particle could approach the nucleus if there were no forces between them is known as impact parameter (b), for head on collision, b =0.

3. Distance of closest approach (D) : It's the minimum distance between the \(\alpha\) -particles. For head on collision and target nucleus at which initial K.E of\(\alpha\) -particles is converted into P.E between\(\alpha\) -particles and target nucleus.

Let us consider a \(\alpha\) -particle at a great distance from the nucleus is approaching a head-on collision with a kinectic energy K. At a closest distance D at point p as shown in figure above the repulsive force of the nucleus stops the approaching \(\alpha\)-particle momentarily and all K.E transforred into P.E. So, we have,$$K=\frac{1}{4\pi\epsilon_\circ}\frac{2e.Ze}D$$$$or\;=\frac{1}{4\pi\epsilon_\circ}\frac{2Ze^2}{K}$$

Relation between b and \(\theta\):

Figure

Consider an\(\alpha\) -particles of mass 'm' is incident on the target nucleus having charge +Ze, where Z is atomic number of target atom. Since the interaction between\(\alpha\) -particle and target nucleus is elastic collision, the K.E remains conserved.

Intitial K.E of \(\alpha\) particles = Finel K.E of scattered \(\alpha\) particle

$$\frac{P_1^2}{2M}=\frac{p_2^2}{2M}$$ Where \(P_1\) and \(P_2\) be the initial and finel momentum of \(\alpha\) -particle.$$P_1=P_2=MV\dotsm(3)$$The change in linear momentum of\(\alpha\) -particles is given by,$$\Delta\overrightarrow{P}=\overrightarrow{P_2}-\overrightarrow{P_2}$$$$=\int_{t=0}^\infty\overrightarrow{F}.dt$$$$or\;\Delta{P}=\int_0^\infty{F}\cos\phi.dt$$$$\Delta{P}=\int_o^\infty\frac{1}{4\pi\epsilon_\circ}\frac{2Ze^2}{r^2}\cos\phi.dt$$$$=\int_{-(\frac{\pi-\theta}{2})}^{(\frac{\pi-\theta}{2})}\frac{1}{4\pi\epsilon_\circ}\frac{(2Ze^2)}{r^2}\cos\Phi\frac{dt}{d\phi}d\phi$$$$=\frac{1}{4\pi\epsilon_\circ}.2Ze^2\int_{-(\frac{\pi-\theta}{2})}^{(\frac{\pi-\theta}{2})}\frac{\cos\phi}{r^2}\frac{dt}{d\phi}d\phi\dotsb(4)$$The force between incident \(alpha\) -particle and target nucleus is central force. Inside of central force field angular momentum remains conserved. i.e $$L=Mvb=Mr^2\omega$$$$or\;vb=r^2\frac{d\phi}{dt}$$$$or\;\frac{dt}{d\phi}=\frac{r^2}{vb}\dotsm(5)$$From \(eq^n\) (4) and (5) we get,

$$\Delta{P}=\frac{2Ze^2}{4\pi\epsilon_\circ}\cdot\frac{1}{vb}\int_{-(\frac{\pi-\theta}{2})}^{(\frac{\pi-\theta}{2})}\frac{\cos\phi}{r^2}\cdot\frac{r^2}{vb}d\phi$$$$=\frac{2Ze^2}{4\pi\epsilon_\circ}\frac{1}{vb}[\sin\phi]_{(\frac{\pi}{2}-\theta)}^{(\frac{\pi-\theta}{2})}$$$$=\frac{2Ze^2}{4\pi\epsilon_\circ}\frac{1}{vb}\biggl[\sin\biggl(\frac{\pi}{2}-\frac{\theta}{2}\biggr)+\sin\biggl(\frac{\pi}{2}-\frac{\theta}{2}\biggr)\biggr]$$$$=\frac{2Ze^2}{4\pi\epsilon_\circ}\frac{1}{vb}2\cos\frac{\theta}{2}$$$$=\frac{4Ze^2}{4\pi\epsilon_{\circ}vb}\cdot\cos\frac{\theta}{2}\dotsm(6)$$

From figur using sin law

$$\frac{\Delta{P}}{\sin\theta}=\frac{P_2}{\sin(\frac{\pi-\theta}{2})}$$$$\Delta{P}=\frac{\sin\theta.P_2}{\sin\frac{\pi-\theta}{2}}\;\;\;\;\; \;i.e \;\; (sin2\theta= 2\sin\theta.\cos\theta)$$

$$\Delta{P}=2MV\sin\frac{\theta}{2}\dotsm(7)$$

From equation (6) and (7)

$$2MV\sin\frac{\theta}{2}=\frac{4Ze^2}{4\pi\epsilon_{\circ}vb}\cos\frac{\theta}{2}$$$$\frac12MV^2=\frac{Ze^2}{4\pi\epsilon_\circ}\frac{\cot\frac{\theta}{2}}{b}$$$$or\;K= \frac{Ze^2}{4\pi\epsilon_circ}\frac{cot\frac{\theta}{2}}{b}$$$$b=\frac{4Ze^2}{4\pi\epsilon_\circ}\frac{\cot\frac{\theta}{2}}{K}\dotsm(8)$$Where\(K=\frac{1}{4\pi\epsilon_circ}\frac{2Ze^2}{D}\dotsm(9)\)

Using equation (8) in equation (9) we get, $$b=\frac{Ze^2}{4\pi\epsilon_\circ}\frac{\cot\frac{\theta}{2}}{\frac{1}{4\pi\epsilon_\circ}\frac{2Ze^2}{D}}$$

Which is the required relation between impact parameter, distance of closest approach and angle of scattering.

Limitation of Rutherford's atomic model

1. According to Rutherford's atomic model an electorn continously revolving around the nucleous in circular orbits but according to electromagnetic theory any charged particle revolving will emit radiation and lose energy so that it should lose energy and fell into nucleus this causes atom collapse. But this not happened, atom is very much stable. This is one drawback of rutherford's atomic model.

2. This theory doesn't explain anything about the arrangement of electrons.

Reference:

Reviews of Modern Physics. Lancaster, P.A.: Published for the American Physical Society by the American Institute of Physics, 1952. Print.

Wehr, M. Russell, and James A. Richards. Physics of the Atom. Reading, MA: Addison-Wesley Pub., 1984. Print.

Young, Hugh D., and Roger A. Freedman. University Physics. Boston, MA: Pearson Custom, 2008. Print.

Adhikari, P.B. A Textbook of Physics. 2070 ed. Vol. II. Kathmandu: Sukunda Publication, 2070. Print.