nuclear physics

Nuclear Physics

The famous Geiger-Marsden Alpha scattering experiment (under Rutherford’s guidance)

In 1909, Geiger and Marsden were studying how alpha particles are scattered by a thin gold foil.

Alpha source

Thin gold foil

Geiger-Marsden

As expected, most alpha particles were detected at very small scattering angles

Alpha particles

Thin gold foil Small-angle scattering

Geiger-Marsden

To their great surprise, they found that some alpha particles (1 in 20 000) had very large scattering angles

Alpha particles

Thin gold foil Small-angle scattering

Large-angle scattering

Explaining Geiger and Marsdens’ results

The results suggested that the positive (repulsive) charge must be concentrated at the centre of the atom. Most alpha particles do not pass close to this so pass undisturbed, only alpha particles passing very close to this small nucleus get repelled backwards (the nucleus must also be very massive for this to happen).

Remember on this scale, if the nucleus is 2 cm wide, the atom would be 200 m wide!

Rutherford did the calculations!

Rutherford calculated theoretically the number of alpha particles that should be scattered at different angles (using Coulomb’s law). He found agreement with the experimental results if he assumed the atomic nucleus was confined to a diameter of about 10-15 metres.

Closest approach

Using the idea of energy conservation, it is possible to calculate the closest an alpha particle could get to the nucleus during a head-on collision.

Alpha particle

nucleus

Closest approach

Initially, the alpha particle has kinetic energy = ½mu2

K.E. = ½mu2

Closest approach

At the point of closest approach, the particle reaches a distance b from the nucleus and comes momentarily to rest.

b

K.E. = 0

Closest approach

All the initial kinetic energy has been transformed to electrical potential energy.

K.E. = 0

b

Closest approach

Using the formula for electrical potential energy which is derived from Coulomb’s law

Kinetic energy lost = Electrical potential

½mu2 = 1 q1q2

4πεo b

K.E. = 0

b

Closest approach

Rearranging we get;

b = 1 q1q2

4πεo ½mu2

K.E. = 0

b

Closest approach

For an alpha particle, m = 6.7 x 10-27 kg, q1 = 2 x (1.6 x 10-19 C) and u is around 2 x 107 m.s-1. If the foil is made of gold, q2 is 79 x (1.6 x 10-19 C).

b = 1 q1q2

4πεo ½mu2

b = 1 x (2 x 1.6 x 10-19 C) x (79 x 1.6 x 10-19 C)

4π x 8.854 x 10-12 Fm-1 ½ x 6.7 x 10-27 kg x (2 x 107 m.s-1)2

b = 2.7 x 10-14 m

The mass spectrometer

A VERY useful machine for measuring the masses of atoms (ions) and their relative abundances. What is it and how

does it work Mr Porter?

The Mass Spectrometer

ions produced

velocity selector

ions accelerated

Region of magnetic field

detector

ion beam

The Mass Spectrometer

Electrons are produced at a hot cathode and accelerated by an electric field. Collision with gas particles produces ions.

Substance to be tested is turned into a gas by heating.

The Mass Spectrometer

Ions acclerated by an electric field

The Mass Spectrometer

Ions enter the velocity selector which contains an electric field (up the page) and magnetic field (into the page) at right angles to each other. By choosing a suitable value for the magnetic field the ions continue in a straight line. i.e. The force produced by the electric filed (eE) is equal to the force produced by the magnetic field (Bev). eE = Bev

The Mass Spectrometer

eE = Bev

or v = E/B

This means that only ions with a specific velocity pass through this region. (Hence ”velocity selector”)

The Mass Spectrometer

The selected ions all with the same velocity (but different masses of course) enter the second region of magnetic field (also into the page). They are deflected in a circular path.

The Mass Spectrometer

Heavier ions continue forward and hit the sides, as do ions that are too light. Only ions of one particluar mass reach the detector.

The radius of the circle is given by R = mv/eB so the mass can be calculated from

m = ReB/v

The Mass Spectrometer

The magnetic field can be varied so that ions of different mass can be detected (higher B would mean that ions of larger mass could be directed at the detector).

The Mass Spectrometer

The detector can measure the numbers of ions detected, hence giving an idea of relative abundance of different ions.

The Mass Spectrometer

The mass spectrometer is particulary useful for identifying isotopes of the same element.

Nuclear energy levels

We have seen previously that electrons exist in specific energy levels around the atom.

There is evidence that energy levels exist inside the nucleus too.

Wow!

Nuclear energy levels

When a nucleus decays by emitting an alpha particle or a gamma ray, the particles or photons emitted are only at specific energies (there is not a complete range of energies emitted, only certain specific levels).

Nuclear energy levels

An alpha particle or photon thus has an energy equal to the difference between energy levels of the nucleus.

energy levels in 235U (MeV)

51.57

0.051

0.013

0.000

Nuclear energy levels

In the alpha decay of 239Pu to 235U, the plutonium nucleus with an energy of 51.57 MeV can decay into Uranium at 3 different energy levels.

energy levels in 235U (MeV)

51.57

0.051

0.013

0.000

Plutonium-239

Nuclear energy levels

If the 239Pu (51.57 MeV) decays to the ground state of 235U (0 MeV), an alpha particle of energy 51.57 MeV is emitted.

energy levels in 235U (MeV)

51.57

0.051

0.013

0.000

Plutonium-239

Alpha emission (51.57 MeV)

Nuclear energy levels

If the 239Pu (51.57 MeV) decays to the 2nd excited state of 235U (0.051 MeV), an alpha particle of energy 51.57-0.051 = 51.52 MeV is emitted. The uranium nucleus is now in an excited state so can decay further by gamma emission to the ground state.

energy levels in 235U (MeV)

51.57

0.051

0.013

0.000

Plutonium-239

Alpha emission (51.52 MeV)

Gamma emission (0.051 MeV)

Nuclear energy levels

In fact the nucleus could decay first to the 0.013 level, and then the ground state, thus emitting two gamma photons.

energy levels in 235U (MeV)

51.57

0.051

0.013

0.000

Plutonium-239

Alpha emission (51.52 MeV)

Gamma emission (0.038 MeV)

Gamma emission (0.013MeV)

Radioactive decay

Beta (β) and positron decay (β+) decay

Positron? That sounds interesting, what is it?

Radioactive decay

In beta decay, a neutron in the nucleus decays into a proton, an electron and an antineutrino.

n p + e + ve 10

11

0-1

00

Radioactive decay

In positron decay, a proton in the nucleus decays into a neutron, a positron (the antiparticle of the electron) and a neutrino.

p n + e + ve 11

10

0+1

00

Radioactive decay

We must not think that the decaying particle actually consists of the three particles in which it splits.

The antineutrino

The antineutrino in beta decay was not detected until 1953, although its presence had been predicted theoretically.

n p + e + ve 10

11

0-1

00

The antineutrino

The mass of the neutron is bigger than that of the proton and electron together.

n p + e + ve

1.008665u – (1.007276 + 0.0005486)u = 0.00084u

10

11

0-1

00

1.008665 u 1.007276 u 0.0005486 u

The antineutrino

This corresponds (using E = mc2) to an energy of 0.783 MeV.

n p + e + ve 10

11

0-1

00

The antineutrino

This extra energy should show up as kinetic energy of the products (proton and electron). Since the electron should carry most of the kinetic energy away, so we should observe electrons with an energy of about 0.783 MeV.

n p + e + ve

In fact we observe electrons with a range of energies from zero up to 0.783 MeV.

10

11

0-1

00

The antineutrino

Where is the missing energy? In 1933 Wolfgang Pauli and Enrico Fermi hypothesized the existence of a third very light particle produced during the decay. Enrico Fermi coined the term neutrino for the ”little neutral one”

n p + e + ve 10

11

0-1

00

Ahhhhh! The little neutral one!

The radioactive decay law

If the number of nuclei present in a sample at t = 0 is N0, the number N still present at time t later is given by

N = Noe-λt

where λ is the decay constant (the probability that a nucleus will decay in unit time)

The radioactive decay law

N = Noe-λt

No

Number of original nuclei present

time

Half-life

After one half-life, the number of original nuclei present is equal to N0/2. Putting this into the radioactive decay law;

N0/2 = N0e(-λt½) where t½ is the half-life

Half-life

N0/2 = N0e(-λt½)

taking logarithms we find

λt½

= ln2

λt½

= 0.693

This is the relationship between the decay constant and the half-life.

Measuring half-life

For short half-lives, the half life can usually be measured directly.

Measuring half-life

For longer half lifes, values of activity can be measured and the decay law can be used to calculate λ and thus t½.

Measure the activity A and chemically find the number of atoms of the isotope.

Use A = λN and then λt½ = ln2

Questions!

• Page 412 Questions 1, 2, 3.

• Page 413 Questions 6, 7, 8, 10.