Question

In order to investigate the structure of atoms, Ernest Rutherford performed his famous experiment, in which he bombarded gold atoms with alpha particles and studied the scattering of the alpha particles. Imagine that an alpha particle (a helium nucleus, consisting of two protons and two neutrons) is initially moving along the x-axis in the positive direction straight toward an initially stationary gold nucleus (containing 79 protons and 118 neutrons) and all subsequent motion takes place along the x-axis. The alpha particle starts with kinetic energy of 8.9 MeV (= 8.9 × 10⁶ eV) far from the gold nucleus. Take the mass of a nucleon (a proton or a neutron) to be 1.7 × 10^-27 kg and assume that the mass of a nucleus equals the sum of the masses of its constituent nucleons. Assume also that all speeds are low compared to the speed of light.

a Find the final momentum of the alpha particle (with its sign), long after it interacts with the gold nucleus, in units of kg⋅m/s.

b. Find the final momentum of the gold nucleus (with its sign), long after in interacts with the alpha particle, in units of kg⋅m/s.

Answer 1

The bombardment of gold atoms with alpha particles will be a perfectly elastic collision. In an elastic collision kinetic energy of the colliding particles are conserved. Also in any collision momentum of the colliding particles are also conserved .We are using these two concepts here.

Given data

• Mass of alpha particle m1=4× 1.7 ×10-27 kg = 6.8 ×10-27 kg

• Mass of the gold nucleus which was at rest initially m2=(79+118) × 1.7 ×10-27 kg =197× 1.7 ×10-27 kg =3.35 ×10-25 kg
• Initial Kinetic energy of the alpha particle K1=8.9×106 eV= 1.426×10-12 J

Let P1 be Initial momentum of the alpha particle.

v1,  v2,   be the velocity of the alpha particle before and after collision and v3 be the velocity of gold nucleus.

Assume that after the collision, the alpha particle bounce back and the gold nucleus moves in the positive x direction.

Initial momentum of the alpha particle   =   

      kg.m/s

Also P1 = m1v1

Therefore ,initial velocity of alpha particle v1 = P1 / m1

   m/s

Applying the law of conservation of momentum

m1v1 = −m1v2 + m2v3

⟹m1×(v1+v2)=m2v3 ----------------------------------------- eqn 1

from the law of conservation of kinetic energy ,

---------------------------------------eqn 2

On dividing the second equation by first equation we get

Substituting the above value of v3 in eqn 1, we get

   and its direction will be negative.

(a)  final momentum of the alpha particle

  

(b) Velocity of the gold nucleus after the collision,

  

  

final momentum of the gold nucleus