In this particular article Derivation for Rutherford scattering formula, we are going to discuss the relationship between impact parameter (b) and (θ). we will also discuss the Rutherford scattering formula in detail along with some basic knowledge of scattering cross-section area.

The relation between impact parameter (b) and the scattering angle (θ)

According to the Conservation law of angular momentum, Angular momentum remains to conserve.

Angular momentum at P,



v/v₀=b/x (2)

From the energy conservation law,

Total energy at point P = Total energy at point A

kinetic energy + potential energy= kinetic energy+ potential energy




x= 2Ze²/4πε₀[1-v²/v₀²]⁻¹ (3)

By the property of hyperbola we have,

NO=ε× OA

Where ε is eccentricity.

And the value of ε is secφ

NO=secφ(OA) (4)

From diagram we have,

x= AO+ON

From equation (3)




Similarly From diagram


x= (1+cosφ)b/sinφ (5)

Put this value in(2),

v/v₀=sinφ/(1+cosφ) (6)

Now put the value of v/v₀ And x in equation (3)

(1+cosφ) b/sinφ=4Ze²/4πε₀ mv₀²[1-sin²φ/(1+cosφ)²]⁻¹

b= 4Ze²/4πε₀ mv₀²(sinφ/(1+cosφ)[1-sin²φ/(1+cosφ)²]⁻¹

now,b= 4Ze²/4πε₀ mv₀²(sinφ/(1+cosφ)[((1+cosφ)²−sin²φ)/(1+cosφ)²]⁻¹

and,b= 4Ze²/4πε₀ mv₀²(sinφ/(1+cosφ)[((1+cosφ²+2 cosφ)−sin²φ)/(1+cosφ)²]⁻¹

again,b= 4Ze²/4πε₀ mv₀²(sinφ/(1+cosφ)[((cosφ²+sinφ²+cosφ²+2 cosφ)−sin²φ)/(1+cosφ)²]⁻¹

also,b= 4Ze²/4πε₀ mv₀²(sinφ/(1+cosφ)[(2cosφ²+2 cosφ)/(1+cosφ)²]⁻¹

b= 4Ze²/4πε₀ mv₀²(sinφ/(1+cosφ)[((1+cosφ)²/2cosφ(1+cosφ)]⁻¹

b=4Ze²/4πε₀ mv₀²(tanφ)

from diagram tanφ=tan/2(π-θ)



b=4Ze²/2(4πε₀) mv₀²(cotθ/2)

and, b=2Ze²(cotθ/2)/2(4πε₀) (1/2mv₀²)

also, b=Ze²cotθ/2/(4πε₀)E

Above equation shows the realtion between b and θ.

Rutherford scattering formula

Let us consider that the α-particles are bombarded on the thin gold foil. The thickness of this gold foil is t. The number of atoms in the gold foil is n per unit volume. It is clear that that the scattering of the alpha particle from the gold foil at the scattering angle greater than θ when impact parameter is zero and b. The transverse is πb² of the cylinder is called scattering cross-section.

The number of gold atoms whose impact parameter is b or less than b is nπb²t. This is also the probability of alpha particles scattered at the angle θ or greater than θ.
Hence probability f= nπb²t            (1)
And we already know that,
b= KZe²(cotθ/2)/E                          (2)
Now put the value of b from Equation (2) into equation (1)
f=nπt(KZe²)²(cot²θ/2)/E²            (3)
Here N is the number of alpha particle striking per unit area per second. Then the number of alpha particles scattered into an angle between θ and θ+dθ is given by-
dN = Ndf
Or dN = nN πt(KZe²)²(cot²θ/2)(cosec²θ/2)dθ/E²


If we consider that a screen is placed between the gold foil such that the distance of the screen and foil is in the direction of scattering angleθ is R. Then the number of dN particles incident on the area (2πR sinθ)(Rdθ) of the screen.

Hence the number of the alpha particles falling per unit area of the screen can be given by-

N(θ)=dN/(2πR sinθ)(Rdθ)



sin θ=2 sinθ/2 (cos θ/2)

∴ N(θ)=nNt(KZe²)² ((cosec²)² θ/2)dθ/4E²R²           (4)

Derivation for Rutherford scattering formula

equation (4) is called Rutherford scattering formula. It is clear from this equation that number of alpha particles scattered into an angle between θ and θ+dθ and falling on a unit area of the screen is proportional to –

(a) Fourth power of cosecant of the half scattering angle

(b)The thickness of the foil

(c)Number of scatterer per unit volume

(d)Number of alpha particles striking per unit area per second

(e)Square of charge in the nucleus of the scatterer and inversely proportional to (1) square of the energy of alpha particles. In 1909 and 1914 Geiger and Marsden proved these results. They observed the variation of N(θ) with θ. According to them, the scattering of nearly 0.14% alpha particles occurs within a scattering angle greater than 1º.

Scattering cross-section

The Scattering cross-section is defined as the ratio of the number of alpha particles scattered per unit solid angle per unit time in a particular direction and the intensity of the particles.

∴Scattering cross section σ(θ,φ)= Number of scattered particle per unit solid angle per unit time/Intensity of incident particles

where the solid angle between θ and θ+dθ is-

dΩ=2π sinθ dθ

The conclusion of Derivation for Rutherford scattering formula

In this perticular article Derivation for Rutherford scattering formula, we have discuss about Derivation for Rutherford scattering formusctterattering cross section etc in easiest way possible.


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