KJM 5900 Exercise 5 - Neutron Activation of Silver

When a particle collides with a nucleus, the collision can either be elastic (billiard ball type of collision) or the particle can be captured, resulting a new nucleus. The outcome of a collision depends on the particles energy and type, and the type of nucleus. If the particle is captured, the new particle will most likely be excited. The new nucleus can deexcite sending out a γ-ray photon or a particle, or both.

Neutron Source

In this lab exercise you will use a neutron source. It consists of beryllium metal (powder) mixed with 450 GBq (12 Ci) of ²³⁸Pu, an alpha-emitter. When the alpha-particles hit the Be nuclei the following nuclear reactions take place:

⁹Be(alpha,n)¹²C

The source emits 3 ?10⁷ neutrons per second. Most often we want to know the number of neutrons that hits our target, the neutron flux. That is, the number of emitted neutrons per unit area and time. The neutrons emitted form the source have an average energy of 4-5 MeV and are called fast neutrons.

Thermal Neutrons

Thermal neutrons have and average energy of 0.025 eV and an energy distribution comparable to that of gas molecules at room temperature. At this low energy the probability of neutron capture is large. So we want to reduce the energy of the fast neutrons from our source to thermal energies.

Since neutrons are neutral they do not lose their energy through electrostatic interactions. Rather they lose energy in collisions with other nuclei. The most efficient transfer of energy is when the colliding neutron and nucleus have the same mass. So, materials containing a lot of hydrogen are good materials for moderating fast neutrons, two examples are paraffin and water.

Thermal neutrons will move in all directions, due to the collisions with the nuclei in the moderator. They can then be considered a gas, filling the moderator, where the density decreases with the distance from the source.

Neutron Activation

Nuclear reactions between a nucleus ^AM and neutrons will usually be the capture of a single neutron:

Usually we write such reactions in the more compressed form

The γ ray is emitted in order for the new nucleus to get rid of excess energy.

The new isotope is of the same element, but the mass number has increased by one. Since the mass number changed, the isotope may be radioactive. (Use your Nuclear Chart and consider what happens when you irradiate e.g. naturally occuring Si with thermal neutrons. Which Si isotopes are produced and which of these are radioactive?)

It is necessary to be able to calculate the amounts which will be produced in neutron activation. This we can do by taking into account

• The flux phi of thermal neutrons.
• The number N_T of nuclei irradiated (T stands for Target).
• The probability of a reaction taking place.
  
• The time t the irradiation lasts.

The probability for capturing a neutron is refered to as the cross section and denoted by sigma.

The production rate F (the number of nuclear reactions per unit time) of the new nucleus is then given by the following formula:

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?(1)

Usually we are interested in cases where the produced nucleus is radioactive. Then we must also take into account how fast this nucleus disapear due to its radioactive decay:

? ? ? ? ? ? ? ? ? ? ? ? (2)

lamda is the decay constant of the new nucleus.

Remember that the production rate matematically is the change in the number of produced nuclei N per unit time: F = dN/dt. Thus, equation (2) is an differential equation which can be solved to give the number of produced nuclei at the end of irradiation:

? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? (3)

Here N₀ is the amount of the nucleus ^A+1M after an irradiation time τ.

Since the activity A is equal to lamda x N we can use equation (3) to calculate how much activity we will have after irradiation:

? ? ? ? ? ? ? ? ? ?(4)

After irradiation the produced nuclei will decay the normal way and we can thus calculate the amount of activity A after a time t since the end of irradiation by:

? ? ? ? ? ? ? ? ? ?(5)

Please notice:

t = time after irradiation
tau? = time of irradiation

T = target

In-Growth Time

When T >> T_1/2 in equation (2) the maximum decay rate possible is A₀ = sigma x phi x N_T. This condition is called saturation.

Time of irradiation	Saturation
0,01T1/2	0,7%
0,1T1/2	7%
0,5T1/2	29%
1,0T1/2	50%
2,0T1/2	75%
3,0T1/2	87,5%
10,0T1/2	99,9%

Φ : neutron flux, cm^-2s_-1

σ : reaction probability, cm² or barn, 1 barn = 10^-24 cm²

N_T: number of atoms in target

D₀ : decay rate at end of irradiation (dpm or Bq)

Silver Isotopes

Naturly occuring silver consist of two isotopes: ¹⁰⁷Ag (51,83%) and ¹⁰⁹Ag (48,17%). Therefore, when irradiating with neutrons we can get two new isotopes, both radioactive: ¹⁰⁸Ag and ¹¹⁰Ag. Since both of these have metastable states, we have four different possible products

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Last updated by Jon Petter Omtvedt at 7. October 2004