Q14.

The diagram below shows a grid of neutron number against proton number. A nucleus is marked in place:

(a) Draw arrows on the grid, each starting on X and ending on a daughter nucleus after the following transitions:

(i) β^– emission (label this arrow A), neutron emission (label this arrow B), electron capture (label this arrow C).

β^– decay will result a neutron changing into a proton - so neutron number goes down and proton number goes up!

Neutron emission will just decrease the number of neutrons

Electron capture will result a proton changing into a neutron - so neutron number increases and proton number goes down!

(ii) Give the equation for electron capture by the nucleus .

+ +

(4 marks)

(b) When decays to by β^– decay, the daughter nucleus is produced in one of two possible excited states. These two states are shown in below together with their corresponding energies.

(i) Calculate the maximum possible kinetic energy, in J, which an emitted β^– particle can have.

The question tells you there are two possible excited states. It tells you the magnesium nucleus emits a beta particle and leaves the daughter nucleus in an excited state.

So, you have to ignore the ground state aluminium value, as it tells you in the question that the beta emission results in an excited Aluminium nucleus.

The energy released as magnesium changes to the lower energy aluminim will result in the KE of the electron and the lower potential energy of the nucleus. If a lower energy state nucleus results, then the beta particle has more kinetic energy and vice versa.

So, the maximum kinetic energy of the beta particle results when the decay produces an aluminium nucleus in the lower energy state after the decay.

The change in energy is (4.18 - 1.33) x 10^-13 = 2.85 x 10^-13 J

(ii) The excited aluminium nuclei emit γ-photons. Calculate each of the three possible  γ-photon energies

See the diagram above. The gamma ray photon is emitted when the metastable state aluminium nuclide goes to ground state. The diagram shows the three possible stages of grounding.

A - full jump (one stage from highest excited state): 1.63 x 10^-13 J

C - full jump (one stage from the lower excited state): 1.33 x 10^-13 J

B - Jump between excited states: (1.63 - 1.33) x 10^-13 J = 0.30 x 10^-13 J (all three must be correct for the one mark)

(iii) Calculate the frequency of the most energetic photon emitted.

E = hf

f = E/h

f = 1.63 x 10^-13/(6.63 x 10^-34)

f = 2.46 × 10²⁰ Hz

(a 'carry-forward error' mark was allowed from (b)(ii) if the largest value was taken)

(3 marks)

(c)

(i) State and explain two precautions that should be taken when working with a sample of in a school laboratory.

They like you to link safety to reducing time of exposure and reducing intensity of exposure.

Handle the sample with (long) (30 cm) tweezers because the radiation intensity decreases with distance (inverse square law of radiation).

Store the sample in a lead box (immediately) when not in use to avoid unnecessary time of exposure to radiation

The precautions must have a reason to get the mark. The question says 'state and explain'.

 

(ii) Discuss which of the two types of radiation, β^– or γ, emitted from a sample of would be the more hazardous.

You could get the mark for either argument! You just had to realise in which way it was more dangerous compared to the other.

γ rays are more penetrating and are therefore more hazardous, as they are able to easily penetrate and ionise the internal organs of the body.

OR

β^– particles are not as penetrating as gamma, but can still penetrate into the body when the source is outside. They are more hazardous than gamma because they are more ionising and damage by them will be more concentrated within the body tissue; thereby increasing the possibility of tissue damage or mutated cells proliferating.

(3 marks)

(Total 10 marks)