Chapter 5 Lecture 4 Rutherford Scattering

Akhlaq Hussain 1 5.5 Rutherford Scattering Formula

Rutherford’s experiment of 훼-particles scattering by the atoms in a thin foil of gold revealed the existence of positively charged nucleus in the atom. ➢ The 훼-particles were scattered in all directions. ➢ Some passes undeflected by the foil. ➢ Some were scattered through larger angles even back scattered. ➢ The only valid explanation of large angle deflection was if the total positive charge were concentrated at some small region. ➢ Rutherford called it nucleus. 2 5.5 Rutherford Scattering Formula

′ 풖ퟏ

3 5.5 Rutherford Scattering Formula

Consider an incident charge particle with charge z1e is scattered through a target with charge z2e. The trajectory of scattered particles is a hyperbola as shown. The trajectory of the particles in repulsive field with origin at o' is shown. In the case of attractive force, center will be at “o”. when particles are apart at large distance, the total energy it posses is in form of Kinetic energy

′ 1 ′ 2 푘 = ൗ2 µ푢1 = 퐸

′ 2퐸 푢1 = µ 4 5.5 Rutherford Scattering Formula

푆 푑푠 We have to find 휎 휃′ = 푠푖푛 휃′ 푑휃′

′ 푙 = µ푢1푆 = 푆 2µ퐸 Where S is the impact parameter. The trajectory is hyperbola, from equation of conic 훼 = 푒 cos 훼 − 1 and for 푟 → ∞ 푟 1 1 ⇒ cos 훼 = = 푒 2퐸푙2 1+ µ푘2

From the figure 2훼 + 휃′ = 휋

휋−휃′ 훼 = 2

휋−휃′ 휃′ Now cot 훼 = cot = tan 5 2 2 5.5 Rutherford Scattering Formula

휃′ cot 훼 = tan 2

휃′ cos 훼 cos 훼 tan = = 2 sin 훼 1−푐표푠2훼

휃′ cos 훼 tan = 2 1 cos 훼 −1 푐표푠2훼

휃′ 1 tan = 2 1 −1 푐표푠2훼

휃′ 1 1 tan = = 2 푒2−1 2퐸푙2 1+ −1 휇푘2

6 5.5 Rutherford Scattering Formula

−1Τ 휃′ 2퐸푙2 2 tan = 2 휇푘2

휃′ 푘 휇 푘 tan = = ′ 2퐸 2 푙 2퐸 푙푢′ Because 푢 = 1 1 µ ′ Since 푙 = µ푢1S

휃′ 푘 ⇒ tan = ′2 2 푆µ푢1 Differentiating with θ' above equation can be written as.

1 휃′ −푘 푑푠 푠푒푐2 = 2 ′2 ′ 2 2 푆 µ푢1 푑휃

2 ′2 푑푠 푆 µ푢1 ⇒ ′ = 휃′ 7 푑휃 2푘 푐표푠2 2 5.5 Rutherford Scattering Formula

Since the differential scattering Cross-Section is given by. 푆 푑푠 휎 휃′ = 푠푖푛 휃′ 푑휃′

2 ‘2 ′ 푆 푆 µ푢1 휎 휃 = 휃′ 휃′ 휃′ 2 푠푖푛 푐표푠 2푘 푐표푠2 2 2 2

3 ‘2 ′ 푆 µ푢1 휎 휃 = 휃′ 휃′ 4푘 푠푖푛 푐표푠3 2 2 휃′ 푘 Since tan = ퟐ ′ 2 푆µ푢1

휃′ 푘 cos ퟐ ⇒ 푠 = 2 휃′ Putting in above equation µ푢′ sin 1 ퟐ 8 5.5 Rutherford Scattering Formula

′ 3휃 2 푘3 푐표푠 µ푢′ 휎 휃′ = 2 . 1 3 ′ 6 휃′ 휃′ 휃′ µ 푢1 푠푖푛3 4푘 푠푖푛 푐표푠3 2 2 2

푘2 1 휎 휃′ = . 2 ′ 4 휃′ µ 푢1 4 푠푖푛4 2

2 ′ 푘 1 휎 휃 = 2 . 휃′ 4퐸 4푠푖푛4 2 2 Since 푘 = 푧1푧2푒 , Therefore,

2 2 2 ′ 푘 1 푧1푧2푒 1 휎 휃 = 2 . 휃′ = 2 . 휃′ 4퐸 4푠푖푛4 4퐸 4푠푖푛4 2 2

9 5.5 Rutherford Scattering Formula

The total cross section in Centre of mass coordinate system will be

′ ′ ′ ′ 휎퐶.푀 휃 = 2휋 න 휎 휃 sin 휃 푑휃

2 2 ′ 푧1푧2푒 1 ′ ′ 휎퐶.푀 휃 = 2휋 ׬ 2 . 휃′ sin 휃 푑휃 4퐸 4푠푖푛4 2

2 2 ′ ′ ′ 푧1푧2푒 1 휃 휃 ′ 휎퐶.푀 휃 = 휋 2 ׬ 휃′ 2 sin cos 푑휃 2퐸 4푠푖푛4 2 2 2

푧2푧2푒4휋 휃′ 휃′ ′휎 휃′ = 1 2 ׬ 푐표푠푒푐3 cos 푑휃 퐶.푀 4퐸2 2 2

10 5.5 Rutherford Scattering Formula

−19 −14 If 푧1 = 2, 푧2 = 79, e = 1.6 × 10 퐶, 퐸 = 7.9푀푒푉 & 푟0 = 10 m. ′ ➢ If we solve the question for 흈푻 휽 we get a divergent results. ➢ The physical reason is that coulomb field, which has infinite range. ➢ Particles with large impact parameters will be deflected through some angle. ′ ➢ Hence the small but finite contribution led us to an infinite value for the 흈푪.푴 휽 . ➢ In case of atoms, the nuclear coulomb field is screened by electrons around the nucleus and results in finite range and finite cross-section. Such field is represented by 1 −푟 푉 = 푒 Τ푎 푟 “푎 is the screening radius.

11 5.5 Rutherford Scattering Formula Distance of Closest approach The distance at which scattered particles turned away from the scattering center.

푘 푧 푧 푒2 푟 = 푎 1 + 푒 and 푎 = = 1 2 1 2퐸 2퐸

2 2 푧1푧2푒 2퐸푙 푟1 = 1 + 1 + 2 2 4 2퐸 푧1 푧2 푒 µ For the smallest distance 푙 must be zero.

푧 푧 푒2 Therefore, 푟 = 푟 = 1 2 1 + 1 1(푚푖푛) 0 2퐸

푧 푧 푒2 푟 = 1 2 1(푚푖푛) 퐸

푧 푧 푒2 퐸 = 1 2 12 푟0 5.6 Examine the scattering Produced by a Repulsive central force 풇 = 풌풓−ퟑ

푘 1−푥 푑푥 휃푠 휎 휃 푑휃 = ∙ , where 푥 is the ratio Τ휋 and E is the energy. 2퐸 푥2 2−푥 2푠푖푛휋푥

푘 3 Solution: The repulsive central force is 푓 = ൗ푟3 = 푘푢 and the differential equation for the orbit is

푑2푢 휇푓 휇푘푢3 + 푢 = − 푢 = − 푑휃2 푙2푢2 푙2푢2

푑2푢 µ푘 ⇒ + 1 + 푢 = 0 푑휃2 푙2

푑2푢 ⇒ + 훾2푢 = 0 푑휃2

µ푘 훾 = 1 + (1) 푙2

13 The solution of this differential equation is 푢 = 퐴 cos 훾휃 + B sin 훾휃 Before the collision particles is at angle 휃 = 휋 and at 푟 = ∞ 푢(휃 = 휋) = 0 ⇒ 퐴 cos 훾휋 + 퐵 cos 훾휋 = 0 ⇒ 퐴 = −퐵 tan 훾휋 (a)

After collision to 푟 = ∞ at angle 휃 = 휃푠 yield the condition

퐴 cos 훾휃푠 + 퐵 sin 훾휃푠 = 0 Putting the value of A ⇒ −퐵 tan 훾휋 cos 훾휃푠 + 퐵 sin 훾휃푆 = 0

⇒ − tan 훾휋 cos 훾휃푠 + sin 훾휃푆 = 0 14 sin 훾휋 ⇒ − cos 훾휃 + sin 훾휃 = 0 cos 훾휋 푠 푠

⇒ − sin 훾휋 cos 훾휃푠 + sin 훾휃푠 cos 훾휋 = 0

⇒ sin 훾 휃푠 − 휋 = 0 −1 ⇒ 훾 휃푠 − 휋 = sin (0) = 휋 휋 휋 ⇒ 훾 = = 휃 휃푠−휋 휋 푠−1 휋 휃 In terms of 푥 = Τ휋 ퟏ 휸 = 풙−ퟏ Putting in equation (1) and squaring both the sides µ푘 1 1 + = 푙2 푥−1 2

15 µ푘 1 ⇒1 + = 2µ퐸푆2 푥−1 2 µ푘 1 ⇒ = − 1 2µ퐸푆2 푥−1 2

푘 1− 푥−1 2 ⇒ = 2퐸푆2 푥−1 2

푘 1−1−푥2+2푥 −푥 푥−2 ⇒ = = 2퐸푆2 푥−1 2 푥−1 2

푘 푥−1 2 푘 ⇒ 푆2 = − = − 푥 − 1 2푥−1 푥 − 2 −1 2퐸 푥 푥−2 2퐸 Differentiating above equation

푘 2 푥−1 푥−1 2 푥−1 2 ⇒ 2S푑푆 = − − − 푑푥 2퐸 푥 푥−2 푥2 푥−2 푥 푥−2 2

16 푘 2푥 푥−1 푥−2 − 푥−1 2 푥−2 −푥 푥−1 2 ⇒ 2S푑푆 = − dx 2퐸 푥2 푥−2 2

푘 2푥 푥−2 − 푥−1 푥−2 −푥 푥−1 ⇒ 2S푑푆 = − 푥 − 1 dx 2퐸 푥2 푥−2 2

푘 2푥2−4푥−푥2+3푥−2−푥2+푥 ⇒ 2S푑푆 = − 푥 − 1 dx 2퐸 푥2 푥−2 2

푘 −2 ⇒ 2S푑푆 = − 1 − 푥 dx 2퐸 푥2 푥−2 2

푘 1−푥 ⇒ S푑푆 = − dx 2퐸 푥2 2−푥 2 No the differential cross section is given by

푆푑푆 휎 휃 푑휃 = 푠푖푛 휃

푘 1−푥 1 ⇒ 휎 휃 푑휃 = dx 2퐸 푥2 2−푥 2 푠푖푛 휋푥 17