San José State University

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 The Derivation of the Rutherford Scattering Formula with Target Recoil

In 1911 Ernest Rutherford published a formula which indicated that the number of particles that would be deflected by an angle θ due to scattering from fixed nuclei is inversely proportional to the fourth power of the sine function of one half the angle of deflection; i.e.,

#### n(θ)Δθ = [κ/sin4(θ/2)] Δθ

where κ is a constant.

It was brilliant analysis in support of brilliant empirical work. Rutherford allowed a beam of alpha particles (helium nuclei) to impinge upon very thin gold foil. The distribution of the deflected alpha particles corresponded to his formula. This result established that the the structure of atoms involved a small dense positively charged nucleus surrounded by the negatively charged electrons. Prior to Rutherford's work the prevalent concept of the structure of atoms was J.J. Thomson's plum pudding model in which the positive and negative charges of atoms were uniformly intermixed.

The following is a derivation of Rutherford's formula taking into account the recoil of the target nuclei. A notational conventional will be used to simplify the presentation: Upper case letters will be used for the characteristics of the target particle and lower case letters for the impinging particle. In this analysis however there is no reason to label one particle a target rather than the other. The analysis will be carried out for the case that the center of mass of the two particles, point O, is motionless. After the analysis in this center-of-mass coordinate system is completed the results can be transferred to so-called laboratory coordinate system.

Let the initial speeds of the two particles be v and V. The velocity of the particle on the right, which shall be arbitrarily called the target particle, is thus −V.

For the center of mass to be stationary it must be that

#### mv + MV = 0 so V = (m/M)v

where m and M are the particle masses.

The two particles are initially on horizontal trajectories that are separated from the center of mass O by distances b and B. The relation of these two distances is

#### B = (m/M)b

The initial angular momenta with respect to the point O are mvb and MVB.

When the particles are at points x and X where the line between them makes an angle of φ with respect to the horizontal their tangential velocities with respect to O are r(dφ/dt) and R(dφ/dt). This means that their angular momenta are mr(dφ/dt)r=mr²(dφ/dt) and MR²(dφ/dt). The conservation of angular momentum requires that

These imply that

#### (dφ/dt) = vb/r² and (dφ/dt) = VB/R²

Since (v/V)(b/B)/(r/R)² = (m/M)(m/M)/(m/M)² = 1, the above two formulas are consistent.

It is convenient to express the above formulas dt=(r²/vb)dφ and dt=(R²/VB))dφ.

The moment arms r and R are related by

#### R = (m/M)r

Let the angle between the asymptotes to the particle trajectories be denoted as θ. The line OD bisects the angle which is the complement of θ. Thus α=(π−θ)/2=π/2−θ/2. The angle β is π less the sum of the angles α and θ.

The component of the initial linear momentum of the particle arriving from the left is mvcos(β) but since β=π/2−θ/2, this momentum component is equal to mvsin(θ/2). By symmetry the final velocity of this particle is also v. The component of the final linear momentum along the line dOD is also of magnitude mvsin(θ/2) but opposite in sign. Thus the change in linear momentum in the direction of the line dOD is 2mvsin(θ/2). For the particle arriving from the right, the target particle, the change in in the component of the linear momentum along the line dOD is 2MVsin(θ/2). Because of the relation of v and V this is equal in magnitude to 2mvsin(θ/2) but opposite in sign.

A fundamentatl principal of mechanics is that the change in linear momentum p in any direction over a time interval is equal to the the force F in that direction integrated over the time interval, the so-called impulse integral; i.e.,

#### Δp = p(t1) − p(t0) = ∫t0t1Fdt

Presume the force between the two particles is electrostatic and their charges are q and Q.

At the point x the force on the impinging particle is equal to JqQ/(r+R)² (where J is the force constant) and its component along the line dOD is this multiplied by cos(φ). Thus the impulse integral is

#### ∫−∞+∞(JqQ/(r+R)²)cos(φ)dt

But from the relation previously found dt=(r²/vb)dφ so the impulse integral can be transformed to

#### ∫−α+α(JqQ/(r+R)²)cos(φ)(r²/vb)dφ which reduces to (JqQ/vb)∫−α+αcos(φ)dφ

The above impulse integral then further reduces to

#### (JqQ/vb)[sin(α)−sin(−α)] = 2(J/vb)sin(α) and since α=π/2−θ/2 this reduces to 2(JqQ/vb)cos(θ/2)

Since the impulse integral must equal the change of momentum

#### 2(JqQ/vb)cos(θ/2) = 2mvsin(θ/2) which reduces to tan(θ/2) = JqQ/(mbv²) or, equivalently cot(θ/2) = mv²b/(JqQ)

The impact parameter b0, the separation distance of the initial trajectories, is b+B which is equal to (1+m/M)b. Thus b=b0/(1+m/M). When this relationship is substituted into the formula

#### cot(θ/2) = mv²b0/[JqQ(1+m/M)]

In the laboratory coordinate system the initial velocity v0 of the impinging particle is v+V and that of the target particle 0. Thus v0=v+V=(1+m/M)v and hence v=v0/(1+m/M). When this relationship is substituted into the previous formula the result is

#### cot(θ/2) = mv0²b0/[JqQ(1+m/M)³] or, equivalently b0 = JqQ(1+m/M)³/(mv0²)cot(θ/2)

This is the relation between the angle of deflection θ and the impact parameter b0. It only differs from the relation derived in the Rutherford analysis in which the target particle is presumed fixed by the term (1+m/M)³ in the denominator.

## The Distribution of the Angle of Deflection Resulting from the Distibution of the Impact Parameter

A variable z has a cumulative probability function P(Z), which is the probability that z will have a value less than or equal to Z, and a probability density function p(z). The two are related by

#### P(Z) = ∫−∞Xp(z)dz and p(Z) = dP(z)/dz

If a variable y is a monotonic function of z, say y=f(z) then there is an inverse function z=g(y) and dz=g'(y)dy. The probability density function for y is given by

#### q(y) = p(g(y))|g'(y)|

where |g'(y)| is the absolute value of the derivative of the function g(y) with respect to its argument.

Suppose the incoming particles are evenly distributed over a beam of radius B0 and that there is only one target nucleus and it is at the center of the beam. The probability that the impact parameter is less than or equal to a value b0 is the area of a circle of radius b0 around the target nucleus relative to the area of the beam, πB0²; i.e.

#### P(b0) = πb0²/πB0² = B0²/B² and therefore the probability density function for b0 is p(b0) = 2b0/B0²

The relationship between the impact parameter b0 and the deflection angle θ was found to be of the form

#### b0 = γcos(θ/2)/sin(θ/2) = γcot(θ/2)

where γ is equal to JqQ(1+m/M)³/(mv0²)

Therefore db0/dθ = [−γ/sin²(θ/2)](1/2) and |db0/dθ| = γ/(2sin²(θ/2)).

Since the probability density function for b0 is p(b0)=2b0/B0², the probability density function of θ, q(θ), is

#### q(θ) = 2[(γcot(θ/2)/B0²](γ/(2sin²(θ/2)); which reduces to q(θ) = γ²[cos(θ/2)/sin³(θ/2)]/B0²

The minimu deflection angle θmin corresponds to the maximum impact parameter B0.

The relations found above are not quite Rutherford scattering formla. His formula is in terms of the cross section for an interaction. The cross section σ for an impact parameter b0 is the area of a circle with radius b0. Let Θ be the deflection angle corresponding to an impact parameter of b0. The probability that the impact parameter is less than or equal to b0 is the same as the probability that the deflection angle will be greater than or equal to Θ. Given the relations previously found then

#### σ(θ≥Θ) = πb0² = πγ²cot²(Θ/2))

Rutherford's formula is in terms of dσ/dΩ where dΩ is the solid angle between Θ and Θ+dΘ. This solid angle is equal to the area of a band of width dΘ and length equal to the perimeter of a circle of radius sin(Θ); i.e., dΩ- 2πsin(Θ)dΘ.

Note that the argument of the sine function is Θ rather than Θ/2. Since sin(2z) is equal to 2sin(z)cos(z) the above solid angle dΩ is equal to 4πsin(Θ/2)cos(Θ/2). Thus

#### dσ/dΩ = (1/(4πsin(Θ/2)cos(Θ/2))(dσ/dΘ)

The derivative dσ/dΘ is given by

#### πγ²(2cot(Θ/2)(-1/sin²)(Θ/2)(1/2) which reduces to −πγ²cos(Θ/2)/sin³(Θ/2)

Therefore, taking the absolute value and cancelling the cos(Θ/2) terms which appear in the numerator and denominator,

#### dσ/dΩ = (1/4)γ²/sin4(Θ/2) or equivalently dσ/dΩ = (1/4)γ²cosec4(Θ/2)

This is the exact form of the Rutherford formula, the only difference being that the parameter γ is JqQ(1+m/M)³/(mv0²) rather than JqQ/(mv0²).

Rutherford used alpha particles of atomic weight 4.002602 impinging upon very thin gold foil. The atomic weight of gold is 196.966543, thus m/M=0.02032 and hence (1+m/M)³=1.0621. Since the Rutherford formula depends upon γ² this means that the adjustment for target recoil for his experiment involved a factor of 1.1282, a relatively small adjustment given the accuracy of the measurements. Furthermore a constant factor leaves the proportional distribution of particle for the various angles of deflection unaffected.

(To be continued.)

Sources:
W.S.C. Williams, Nuclear and Particle Physics, Clarendon Press, Oxford, 1991.

For more on the scattering and diffraction of atomic particles see SCATTERING.