Rutherford Scattering - Motion under Central Forces

8. Rutherford Scattering

8.1 Introduction

From the study of the trajectory of a body moving under the influence of a central force, the dependence of potential energy on distance, or the law of force, can be determined.

From the law of conservation of energy, the total energy of a particle moving under a central force is

\[ \boxed{ E = \frac{1}{2}m\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{2mr^2} + U(r) } \tag{1} \]

where \(J\) is the angular momentum of the particle and is given by

\[ \boxed{ J = mr^2\frac{d\theta}{dt} } \]

Rearranging equation (1),

\[ \boxed{ U(r) = E - \frac{1}{2}m\left(\frac{dr}{dt}\right)^2 - \frac{J^2}{2mr^2} } \tag{2} \]

To express this equation in terms of the trajectory, let

\[ \boxed{ u = \frac{1}{r} } \]

Since

\[ r = \frac{1}{u}, \]

differentiation with respect to time gives

\[ \frac{dr}{dt} = \frac{dr}{d\theta} \frac{d\theta}{dt}. \]

Since

\[ \frac{d\theta}{dt} = \frac{J}{mr^2}, \]

and

\[ \frac{dr}{d\theta} = -\frac{1}{u^2}\frac{du}{d\theta}, \]

we obtain

\[ \frac{dr}{dt} = -\frac{J}{m}\frac{du}{d\theta}. \]

Therefore,

\[ \left(\frac{dr}{dt}\right)^2 = \frac{J^2}{m^2} \left(\frac{du}{d\theta}\right)^2. \]

Substituting these relations into equation (2), we get

\[ \boxed{ U = E - \frac{J^2}{2m} \left[ \left(\frac{du}{d\theta}\right)^2 + u^2 \right] } \tag{3}

Thus, if the equation of the orbit \(u=u(\theta)\) is known, the dependence of potential energy on the distance \(r\) can be obtained.

8.2 Scattering Method

For atomic particles, the trajectory cannot generally be observed directly because of their extremely small dimensions. Therefore, the potential energy cannot be determined directly from the observed trajectory by the above method.

In such cases, a method known as the scattering method is used to study the interaction between particles.

Scattering: When a moving particle approaches a stationary target particle and is deflected from its original direction due to their interaction, the process is called scattering.

When a light particle is far away from a heavy particle, it moves approximately in a straight line. As it approaches the heavy particle, it enters the interaction region and its path is deflected by the central force.

For a repulsive interaction, the particle approaches the target up to a minimum distance called the distance of closest approach, after which it moves away from the target.

The difference between the initial and final directions of motion is called the angle of scattering, denoted by \(\phi\).

The scattering angle depends on the kinetic energy of the incident particle and the nature of the interaction between the incident and target particles.

8.3 Rutherford's Nuclear Scattering Experiment

Rutherford used the scattering of alpha particles to investigate the internal structure of the atom. The alpha particle carries a charge \(+2e\).

Rutherford showed that the large-angle scattering of alpha particles from a gold foil can be explained by the repulsive Coulomb interaction between the positively charged alpha particle and the positively charged nucleus.

This led to the conclusion that the positive charge and most of the mass of an atom are concentrated in a very small region at its centre, called the nucleus.

Important scale: The atomic diameter is of the order of \(10^{-10}\,\text{m}\), whereas the nuclear size is of the order of \(10^{-14}\,\text{m}\).

8.4 Postulates of Rutherford Scattering Theory

1. The positive charge of an atom is concentrated in a comparatively small region at the centre of the atom. This region is called the nucleus.
2. The mass of the target atom is assumed to be very large compared with the mass of the incident alpha particle. Therefore, the position of the target nucleus can be considered fixed.
3. The scattering is assumed to be due to the repulsive Coulomb interaction between the positively charged alpha particle and the positively charged nucleus. The effect of atomic electrons is neglected.
4. The nucleus and the alpha particle are treated as point charges for the purpose of calculating the scattering trajectory.

8.5 Coulomb Force between Alpha Particle and Nucleus

Let the atomic number of the scattering atom be \(Z\). The charge of the nucleus is therefore

\[ Q_N = +Ze. \]

The charge of an alpha particle is

\[ Q_\alpha = +2e. \]

According to Coulomb's law, the repulsive force between the nucleus and the alpha particle is

\[ F = \frac{K(Ze)(2e)}{r^2}. \]

Therefore,

\[ \boxed{ F = \frac{k}{r^2} } \tag{4} \]

where

\[ \boxed{ k = 2KZe^2 } \]

and

\[ K = \frac{1}{4\pi\varepsilon_0} \approx 9\times10^9\, \text{N m}^2\text{C}^{-2}. \]

Here \(r\) is the distance between the alpha particle and the nucleus.

Potential Energy

For a repulsive Coulomb force,

\[ F = \frac{k}{r^2}. \]

Taking the potential energy to be zero at infinity,

\[ U(r) = -\int_{\infty}^{r} F\,dr. \]

Hence,

\[ U(r) = -\int_{\infty}^{r} \frac{k}{r^2}\,dr. \]

Therefore,

\[ \boxed{ U(r)=\frac{k}{r} } \tag{5} \]

8.6 Trajectory of the Alpha Particle

Let an alpha particle of mass \(m\) approach the nucleus with an initial velocity \(v_0\) from infinity.

At infinity, the potential energy is zero. Hence the total energy of the alpha particle is purely kinetic:

\[ \boxed{ E=\frac{1}{2}mv_0^2 } \tag{6} \]

Let \(b\) be the perpendicular distance between the initial direction of motion and the centre of the nucleus. This distance is called the impact parameter.

Impact parameter \(b\): The perpendicular distance between the initial straight-line path of the incident particle and the centre of the scattering nucleus.

The angular momentum of the alpha particle about the nucleus is

\[ J=mv_0b. \]

Using equation (6),

\[ v_0=\sqrt{\frac{2E}{m}}, \]

and therefore

\[ \boxed{ J=b\sqrt{2mE} } \tag{7} \]

Orbit Equation

From equation (3), with

\[ U=ku, \qquad u=\frac{1}{r}, \]

we obtain

\[ ku = E - \frac{J^2}{2m} \left[ \left(\frac{du}{d\theta}\right)^2 + u^2 \right]. \]

Rearranging,

\[ \boxed{ \left(\frac{du}{d\theta}\right)^2+u^2 = \frac{2m}{J^2}(E-ku) } \tag{8}

Differentiating equation (8) with respect to \(\theta\),

\[ 2\frac{du}{d\theta} \left( \frac{d^2u}{d\theta^2}+u \right) = -\frac{2mk}{J^2}\frac{du}{d\theta}. \]

For \(du/d\theta\neq0\), this gives

\[ \frac{d^2u}{d\theta^2}+u = -\frac{mk}{J^2}. \]

Thus,

\[ \boxed{ \frac{d^2u}{d\theta^2}+u = -\frac{mk}{J^2} } \tag{9} \]

The solution of this differential equation is

\[ u = A\cos\theta - \frac{mk}{J^2}. \]

Defining

\[ p=\frac{J^2}{mk}, \]

the orbit equation becomes

\[ \boxed{ r= \frac{p}{\varepsilon\cos\theta-1} } \tag{10} \]

where \(\varepsilon\) is the eccentricity of the trajectory.

For the present repulsive Coulomb interaction, the trajectory is a hyperbola, for which

\[ \boxed{ \varepsilon>1 } \]

Therefore, the trajectory of an alpha particle under the influence of a repulsive Coulomb force is a hyperbola.

8.7 Relation between Impact Parameter and Scattering Angle

The asymptotes of the hyperbolic trajectory represent the initial and final directions of the alpha particle.

From equation (10), when \(r\rightarrow\infty\),

\[ \varepsilon\cos\theta-1=0. \]

Therefore,

\[ \boxed{ \cos\theta_0=\frac{1}{\varepsilon} } \tag{11} \]

If \(\phi\) is the angle of scattering, geometry gives

\[ \phi=\pi-2\theta_0. \]

Hence,

\[ \frac{\phi}{2} = \frac{\pi}{2}-\theta_0. \]

Therefore,

\[ \cot\frac{\phi}{2} = \tan\theta_0. \]

From equation (11),

\[ \tan\theta_0 = \sqrt{\varepsilon^2-1}. \]

For a Coulomb trajectory,

\[ \varepsilon^2-1 = \frac{2EJ^2}{mk^2}. \]

Using

\[ J^2=2mEb^2, \]

we get

\[ \varepsilon^2-1 = \frac{4E^2b^2}{k^2}. \]

Thus,

\[ \sqrt{\varepsilon^2-1} = \frac{2Eb}{k}. \]

Hence,

\[ \boxed{ \cot\frac{\phi}{2} = \frac{2Eb}{k} } \tag{12} \]

Therefore, the impact parameter is

\[ \boxed{ b = \frac{k}{2E} \cot\frac{\phi}{2} } \]

Since

\[ k=2KZe^2, \]

we obtain

\[ \boxed{ b = \frac{KZe^2}{E} \cot\frac{\phi}{2} } \]

Using \(E=\frac12mv_0^2\),

\[ \boxed{ b = \frac{2KZe^2}{mv_0^2} \cot\frac{\phi}{2} } \]

or equivalently,

\[ \boxed{ \cot\frac{\phi}{2} = \frac{mv_0^2b}{2KZe^2} } \tag{13} \]

Equation (13) gives the relation between the impact parameter and the angle of scattering.

For fixed \(m\), \(v_0\), and \(Z\), a decrease in the impact parameter \(b\) produces an increase in the scattering angle \(\phi\). For a direct collision, \(b=0\), and hence \(\phi=180^\circ\).

8.8 Scattering of Alpha Particles

Consider a gold foil of thickness \(t\), containing \(n\) scattering atoms per unit volume.

According to equation (13), alpha particles scattered through an angle greater than or equal to \(\phi\) correspond to impact parameters between \(0\) and \(b\).

Therefore, the relevant scattering atoms lie inside a cylindrical region of radius \(b\) and thickness \(t\).

\[ \text{Volume of cylindrical region} = \pi b^2t. \]

Since \(n\) is the number of scattering atoms per unit volume, the number of atoms in this region is

\[ N_{\text{atoms}} = n\pi b^2t. \]

Thus, the fraction of alpha particles scattered through an angle greater than or equal to \(\phi\) is proportional to

\[ f=n\pi b^2t. \]

Substituting equation (12),

\[ b = \frac{k}{2E} \cot\frac{\phi}{2}, \]

we obtain

\[ f = n\pi t \left( \frac{k}{2E} \right)^2 \cot^2\frac{\phi}{2}. \]

Therefore,

\[ \boxed{ f = \frac{\pi ntk^2}{4E^2} \cot^2\frac{\phi}{2} } \tag{14} \]

Number of Particles Scattered between \(\phi\) and \(\phi+d\phi\)

Differentiating equation (14),

\[ df = \frac{\pi ntk^2}{4E^2} \frac{d}{d\phi} \left( \cot^2\frac{\phi}{2} \right)d\phi. \]

Since

\[ \frac{d}{d\phi} \left( \cot^2\frac{\phi}{2} \right) = -\cot\frac{\phi}{2} \csc^2\frac{\phi}{2}, \]

the magnitude of the number fraction is

\[ \boxed{ |df| = \frac{\pi ntk^2}{4E^2} \cot\frac{\phi}{2} \csc^2\frac{\phi}{2} \,d\phi } \tag{15} \]

If \(N\) is the number of alpha particles incident per unit area per unit time, then the number of alpha particles scattered between \(\phi\) and \(\phi+d\phi\) is

\[ dN=N|df|. \]

Hence,

\[ \boxed{ dN = \frac{\pi Nntk^2}{4E^2} \cot\frac{\phi}{2} \csc^2\frac{\phi}{2} \,d\phi } \tag{16} \]

8.9 Rutherford Scattering Formula

Suppose a screen is placed at a distance \(R\) from the gold foil. Consider particles scattered between \(\phi\) and \(\phi+d\phi\).

The area of the corresponding ring on the screen is

\[ dA = (2\pi R\sin\phi)(R\,d\phi). \]

Therefore,

\[ dA = 2\pi R^2\sin\phi\,d\phi. \]

The number of particles falling per unit area of the screen is

\[ \frac{dN'}{dA} = \frac{dN} {2\pi R^2\sin\phi\,d\phi}. \]

Substituting equation (16),

\[ \frac{dN'}{dA} = \frac{Nntk^2} {8R^2E^2} \frac{ \cot(\phi/2)\csc^2(\phi/2) }{\sin\phi}. \]

Using the identity

\[ \sin\phi = 2\sin\frac{\phi}{2} \cos\frac{\phi}{2}, \]

and

\[ \cot\frac{\phi}{2} = \frac{\cos(\phi/2)} {\sin(\phi/2)}, \]

we obtain

\[ \boxed{ \frac{dN'}{dA} = \frac{Nntk^2} {16R^2E^2} \csc^4\frac{\phi}{2} } \tag{17} \]

Rutherford Scattering Formula

\[ \boxed{ \frac{dN'}{dA} = \frac{Nntk^2} {16R^2E^2} \csc^4\frac{\phi}{2} } \]

This equation is known as the Rutherford scattering formula.

8.10 Dependence of Rutherford Scattering

From the Rutherford scattering formula,

\[ \frac{dN'}{dA} \propto \frac{Nntk^2}{R^2E^2} \csc^4\frac{\phi}{2}. \]

Since

\[ k=2KZe^2, \]

we have

\[ k^2=4K^2Z^2e^4. \]

Therefore,

\[ \boxed{ \frac{dN'}{dA} \propto \frac{ Z^2 }{ E^2 } \csc^4\frac{\phi}{2} } \]

Thus, the number of alpha particles scattered into a given direction depends on the following factors:

1. It is proportional to the fourth power of \(\csc(\phi/2)\): \[ \frac{dN'}{dA} \propto \csc^4\frac{\phi}{2}. \]
2. It is directly proportional to the thickness \(t\) of the foil.
3. It is directly proportional to the number density \(n\) of the scattering atoms.
4. It is directly proportional to the incident alpha-particle flux \(N\).
5. It is proportional to the square of the atomic number \(Z\): \[ \frac{dN'}{dA}\propto Z^2. \]
6. It is inversely proportional to the square of the kinetic energy \(E\) of the alpha particles: \[ \frac{dN'}{dA}\propto\frac{1}{E^2}. \]

8.11 Experimental Verification

The predictions of Rutherford's scattering theory were tested by Geiger and Marsden through their alpha-particle scattering experiments during the period from 1909 to 1914.

The observed angular distribution of scattered alpha particles was consistent with the inverse-square Coulomb interaction and provided important experimental evidence for the nuclear model of the atom.

8.12 Summary

1. Rutherford scattering is based on the repulsive Coulomb interaction between an alpha particle and a positively charged nucleus.
2. The Coulomb force varies as \[ F\propto\frac{1}{r^2}. \]
3. The corresponding potential energy is \[ U=\frac{k}{r}. \]
4. The trajectory of the alpha particle is a hyperbola.
5. The impact parameter and scattering angle are related by \[ \boxed{ \cot\frac{\phi}{2} = \frac{2Eb}{k} }. \]
6. The Rutherford scattering formula is \[ \boxed{ \frac{dN'}{dA} = \frac{Nntk^2} {16R^2E^2} \csc^4\frac{\phi}{2} }. \]
7. The scattering intensity varies as \(Z^2\) and \(1/E^2\), and has a strong angular dependence \(\csc^4(\phi/2)\).