Nuclear Radius

Question 1

Explain the closest approach method

Answer 1

The repulsive force between the nucleus and alpha particle reduces the speed of the particle to 0 momentarily.

At this point KE = EPE

The radius can be found equating the two equations

Question 2

State an advantage of the closest approach method

Answer 2

• Gives a good estimate of the upper limit for a nuclear radius
• Mathematics are very simple
• Particles are scattered only by the protons and not all the nucleons

Question 3

State 2 disadvantages of the closest approach method

Answer 3

• Value will always be an overestimate
• Alpha particles are hadrons and therefore experience the strong nuclear force which is not accounted for
• Gold nucleus will recoil as alpha particles approaches
• Alpha particles have finite size whereas electrons can be treated as point masses
• Difficult to obtain alpha particles that rebound at 180 degrees
• Alpha particles must all have same initial kinetic energy
• Sample must be extremely thin to prevent multiple scattering

Question 4

State the equation for electron diffraction

Answer 4

sin ø = 0.61 x de Broglie’s wavelength / R

Question 5

Why must the de Broglie’s wavelength of an electron be 1fm?

Answer 5

Because the radius of any nucleus is around 1fm, therefore for significant diffraction the wavelength must be similar to the size of the gap

Question 6

State the equation for nuclear radius

Answer 6

R = k x A^1/3

A - nucleon number
R - nuclear radius

Question 7

Show that nuclear density is constant for all nuclei

Answer 7

p = mass / volume
P = A x m(nucleon) / 4/3πR^3
P = A x m / 4/3πk^3A
P = m / 4/3πk^3 (constant)

Question 8

State the equation for De Broglie’s wavelength

Answer 8

h / mv

Question 9

Explain the electron diffraction method

Answer 9

Electrons are fired at a thin sheet of foil, forming a diffraction pattern with a bright central spot and dimmer concentric circles around it.

A graph of intensity against diffraction angle can be used to find angle of first minimum

Use sinø= 0.61 x DBW / R

Question 10

State an advantage of the electron diffraction method

Answer 10

• More accurate than closest approach method
• Gives direct measurement of the radius of a nucleus
• Electrons do not interact with nucleons in the nucleus via SNF

Question 11

State a disadvantage of the electron diffraction approach

Answer 11

• Electrons must be accelerated to very high speeds to minimise DBW and increase resolution
• Electrons can be scattered by nucleons
  Can make first minimum difficult to determine