Nucleus Size Assignment Help

Atomic Physics - Nucleus Size

Nucleus Size

To calculate the nuclear dimensions from the scattering experiment, Rutherford assumed the following:

(i) The atomic nucleus is so heavy that its motion during the impact is disregarded.

(ii) The nucleus and the alpha particle both are taken as point charges having no dimensions.

(iii) The scattering is due to elastic collision between nucleus and α particle.

Suppose an α particle with initial kinetic energy E is directed towards the centre of the nucleus of an atom. On account of Coulomb’s repulsive force between nucleus and alpha particle, kinetic energy of alpha particle goes on decreasing and in turn, electric potential energy of the particle goes on increasing. At a certain distance r0 from the nucleus, K.E. of  α  particle reduces to zero. The particle stops and it cannot go closer to the nucleus. It is repelled by the nucleus and therefore it retraces its path, turning through 180°. Therefore, the distance r0 is known as the distance of closest approach. At this distance, the entire K.E. of  α  particle is converted into electric potential energy.

Mow, charge on  α  particle, q1 = +2e

Charge on nucleus, q2 = +Ze, where Z is the atomic number of material of the foil and +e is charge on a proton.

Electric potential at distance r0 due to the nucleus

= Ze/4∏ε0r0                                                          (1)

Where 1/4∏ε0 = 9 × 109 Nm2 C-2,

Therefore, potential energy of alpha particle at this distance (r0) from the nucleus

= potential × charge

= Ze/4∏ε0r0 × (2e) = Ze (2e)/ 4∏?0r0               (2)

Kinetic energy of alpha particle of mass m moving with velocity v is 

E = ½ mv2                                                               (3)

If we neglect the loss of energy due to interaction of  α  particle with the electrons, then at the distance of closest approach, is K.E. => P.E.

Therefore, ½ mv2 =  Ze (2e)/ 4∏?0r0 

r0 = Ze (2e)/ 4∏?0 (1/2 mv2)                              (4)

obviously, the radius of the nucleus  must be smaller than the calculated value of r0 as an alpha particle cannot touch the periphery of the nucleus on account of strong repulsion.

In the original experiment,

K.E of  α  particle E = ½ mv2 = 7.7 MeV

= 7.7 × 1.6 × 10-13 joule

i.e. E = 1.2 × 10-12 = 79 for gold and

c = 1.6 × 10-19 coulomb.

From (4), r0 = [1/4∏ε0 × (Ze) (2e)]/1/2 mv2

r0 = [9 × 109 × 79 × 2 (1.6 × 10-19)2]/1.2 × 10-12

r0 = 9 × 79 × 2 × 1.6 × 10-29/1.2 × 10-12

r0 = 3.0 × 10-14 m

= 30 fermi.

This value is considerably larger than the sum of radii of gold nucleus and  α  particle. Thus  α  particle reverses its motion without ever actually touching the gold nucleus.

Example: in a head on collision between an alpha particle and gold nucleus, the closest distance of approach is 4 × 10-14 m. calculating the KE of  α  particle.

Solution: here, r0 = 4 × 10-14 m,

Z = 79 (for gold) KE =?

K.E. = P.E. = (Ze) (2e)/ 4∏?0r0

= 9 × 109 × 79 × 2 × (1.6 × 10-19)/(4 × 10-19)2

= 9.1 × 10-13 J. - Physics Help  - Nucleus Size Assignment Help, Nucleus Size Homework Help, Nucleus Size Assignment Tutors, Nucleus Size Solutions, Nucleus Size Answers, Atomic Physics Assignment Tutors

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