1
Solutions
2011 Question 12 (d)
(i) Name a suitable detector.
GM tube (linked with a ratemeter/scaler)/ Solid state detector
(ii) Describe how the reading on the detector may vary as the paper passes by.
The count rate would decrease with increasing paper thickness.
(iii)Why would the radioisotope Am-241, which emits alpha-particles, not be suitable for this process?
The alpha-particles have poor penetrating power so would be easily blocked by the paper.
(iv) Calculate the number of atoms present in a sample of Sr-90 when its activity is 4250 Bq.
The half-life of Sr-90 is 28.78 years.
1 year = 365 days OR 365.25 days
T½ = 0.693 / λ
λ = 0.693/ T1/2
λ = 7.77 × 10-10 s-1 / 7.63 × 10-10 s-1
Activity = dN/dt = (-) λN
4250 = 7.77 × 10-10(N) OR 4250 = 7.63 × 10-10(N)
N = 5.47 × 1012 (atoms) OR N = 5.57 × 1012 (atoms)
2010 Question 12 (b)
(i) Identify the element X.
Krypton
(ii) Calculate the mass difference between the reactants and the products in the reaction
E = mc2
m = (202.5 × 106)(1.6 × 10−19)
(3 × 108)2)
m = 3.6 × 10-28 kg
(iii)What is a chain reaction?
It is a self-sustaining reaction where fission neutrons go on to produce further fission (giving more
neutrons) etc.
(iv) Give one condition necessary for a chain reaction to occur.
The mass of fuel present must exceed the critical mass / at least one of the neutrons released must cause
fission of another nucleus.
(v) Give one environmental impact associated with a nuclear reactor.
Toxic /radioactive waste, exposure to radiation, etc.
2009 Question 12 (d)
(i) Give the structure of an alpha particle.
It is composed of 2 protons and 2 neutrons
(ii) How are the alpha particles produced?
α-decay is produced when the americium (which is radioactive) undergoes radioactive decay.
(iii)Why do these alpha particles not pose a health risk?
They have a very short range so are either contained within the smoke detector itself or just travel a cm
or two through the air.
(iv) Americium-241 has a decay constant of 5.1 × 10–11 s–1.
Calculate its half life in years.
T½ = 0.693 / λ T
½ = 0.693 / 5.1 × 10–11 T½ = 1.36 × 1010 seconds = 430.6 years
(v) Explain why americium-241 does not exist naturally.
{I don’t think this was a fair question and shouldn’t have appeared on the paper}
Its half life is very short (with respect to age of the universe) and because it is not a member of a decay
series it is not produced ‘in nature’ (it is created artificially).
