9 Nuclear decay Answers to exam practice questions

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1 Pages Exam practice questions 1 X-rays are quanta of energy emitted when electrons fall to a lower energy level, and so do not emanate from the nucleus Answer D. 2 Alpha particles, being the most massive (and therefore slowest moving for a given energy) and charged, cause the most ionisation in air Answer A. 3 Alpha particles have a typical range in air of a few centimetres Answer C. An alpha particle is a helium nucleus, 2He. When an alpha particle is emitted, the mass number will decrease by and the atomic number will decrease by 2 Answer D. 5 A beta particle is equivalent to an electron, 1 0 e. When a beta particle is emitted, the mass number will stay the same and the atomic number will increase by 1 Answer B. 6 One day (2h) is equal to 10 half-lives. The fraction of bromine-83 remaining after 1 day will be: (½) 10 = Answer is C. 7 a) Background radiation is the radiation that is constantly present in our environment. Sources include: radioactive elements in the Earth s crust radioactive gases (mainly radon), emitted by the Earth cosmic rays from outer space medical uses, e.g. X-rays nuclear industry trace amounts of radioactivity in our bodies. [3] b) You would need a Geiger-Müller tube connected to a counter or rate-meter. First of all, the background count should be taken. Then the G-M tube would be held about 1 cm above the soil (not to the side of the beaker as the glass would absorb any alpha particles) and the count-rate noted. A sheet of paper should now be inserted between the soil and the G-M tube. [3] If the count is reduced to background, then only α is present. If the count is reduced considerably, then α with some β and/or γ is present. If the count is not reduced, then only β and/or γ is present.

2 To check further whether β and/or γ is present, a 5 mm sheet of aluminium should be inserted. If the count remains virtually the same, then only γ is present. If the count is reduced a bit, then β and γ are present. If the count is reduced virtually to background, only β is present. Tip: A careful, logical description is needed. This is helped by the use of bullet points. 8 a) Ionisation is when one or more electrons are removed from an atom, leaving the atom with a positive charge. b) An alpha particle consists of two protons and two neutrons (the nucleus of a helium atom). It is emitted from the nucleus of an unstable (radioactive) atom. c) i) As the alpha particle travels through the air, it ionises the air molecules. The energy required to do this (i.e. to remove electrons from the air molecules) comes from the alpha particle s kinetic energy, and so it slows down. ii) The alpha particle eventually picks up two electrons, which are always around in the air, and becomes an atom of helium gas. [3] d) i) Isotopes are different forms of the same element (and so have the same atomic/proton number) having different numbers of neutrons and therefore different mass/nucleon numbers. 21 ii) The symbol 95Am shows that the atomic/proton number is 95 and the mass/nucleon number is 21, which means there are (21 95) = 16 neutrons. iii) 21 95Am p e) i) The half-life of a particular isotope is the average time for a given number of radioactive nuclei of that isotope to decay to half that number. [5] ii) t ½ = ln2 = ln = t ½ 60 ( ) s = s s 1 iii) umber of nuclei = g g = nuclei iv) Activity = ( ) s 1 = Bq

3 f) Americium-21 is a suitable source because: it is an alpha emitter, which means that it readily ionises the air and the alpha particles don t travel very far (making it safe) it has a long half-life, so won t need replacing its activity is such that only a very small mass is needed. [3] a) i) 9Pu ii) iii) 60 27Co 23 Mg U 60 28i 23 11a e + υ e R [3] [Total 20 marks] + e υ e b) [3] [Total 10 marks] 10 a) The half-life of a particular isotope is the average time for a given number of radioactive nuclei of that isotope to decay to half that number. b) With all radioactive sources well away from the area in which the experiment is to be conducted, the background count should be taken. This should be done for sufficient time, say two lots of 5 minutes, to get an average value. The G-M tube should be placed close to the layer of protactinium-23 and the count-rate measured at time intervals of, say, 20 s for about 3 minutes. This will give 10 values (including that at t = 0). The background count should then be subtracted from each reading. [3] A graph of either count-rate A against time t (exponential decay) or ln A against t (straight line of negative slope) should be plotted. If A against t is plotted, the half life can be found directly by determining how long it takes for a particular value of A to decay to A/2. This should be done for three different values of A and an average found. If ln A against t is plotted, the gradient is numerically equal to the decay constant. The gradient should be found by drawing a large triangle and the half-life found from t ½ = ln2. c) If the half-life is 28 years, the change in activity over even a few days would be too small to detect, particularly as the decay is random. A small amount of strontium-90, with a half-life of 28 years, would probably mean that the activity was very low and may not be significantly greater than the background count. [Total marks]

4 ln ln 9 uclear decay Answers to exam practice questions 11 a) i) As alpha particles are charged and massive, they readily ionise matter as they pass through it. They will therefore ionise the chemicals making up human cells and cause terrible internal cell damage. ii) Alpha particles are difficult to detect outside the body as they will be absorbed by even a very thin layer of tissue and so will not penetrate as far as the outside of the body. b) i) From = e t P = e t P = t As = ln2 t ½ P = ln 2 t t ½ 365 days = ln days = 1.83 This gives = e 1.83 = anti-ln of 1.83 = 0.16 This means that the % remaining after one year = 16 % Your graph should look like the one below. Tip: Even if you are asked to sketch a graph, relevant data should be plotted as far as possible. For example, in the figure above some points that you might include are plotted the activity at 1, 2, 3 and half-lives, the activity that has just been found for 1 year and the activity after 2 years. You would only be expected to show 3 or values and you would not be required to use graph paper for a sketch. A grid has been included in the figure above merely for information. [3] ii) = ln2 ln2 = t ½ ( ) s = s 1 Activity A = = s = Bq c) Energy generated by 1 μg = s 1 ( ) ev ( ) J ev 1 = J s 1 The power generated by 1 g would be ( ) W 10 W

5 This is equivalent to a very powerful (150 W) filament lamp. If 1 g were to be contained in a small capsule, this magnitude of power would soon melt the capsule. The equations are: 2He 7 6C 6 C 2He 6 C [Total 16 marks] + e υ e Stretch and challenge [Total 6 marks] 13 a) i) The equation d = is based on the concept that the rate of decay is proportional to the dt number of radioactive nuclei, i.e. the more nuclei there are, the more likely it is that one will decay. ii) If there are very large numbers of nuclei, as there usually are in radioactivity, then it is statistically more probable that the above equation, and other mathematical laws, will make calculations valid. iii) d dt = 1 d ln = t t = dt 0 = e t When t is equal to the half life t ½ we have = 2 = e t 1 2 = e ½ 2 = e ½ 2 = e ½ t ½ = ln 2 t ½ = ln 2 b) i) From t ½ = ln 2 it follows that an isotope with a very short half life t ½ will have a very large decay constant and so even a small amount of the isotope will have a detectable activity. Also, an isotope with a short half-life will not stay in the blood flow very long and so will be less harmful to the patient. ii) Fluorine-18 ( 18 ) does not have enough neutrons in the nucleus to provide the necessary F 9 strong force that balances the electromagnetic repulsion between the protons. If it undergoes β + decay by emitting a positron, it decays into 19 9 F, which is stable. 1 1 Effectively 1 p 0 n e (plus an electron neutrino). iii) In PET, the positrons are annihilated when they encounter an electron. As body tissue does not contain positrons, using electrons would not work, as they would not encounter

6 any positrons to annihilate. iv) From t ½ = ln 2 = ln 2 t ½ = From = e t in one hour: = e t = exp[ s 1 ( s)] ( )s = s 1 = e 0.36 = 0.71 = 0.71 If 71% of the nuclei remain, then 29 % will have decayed. v) By conservation of mass-energy ΔE = c 2 Δm The electron and positron each has a mass of kg, so: ΔE = ( m s 1 ) 2 2 ( kg) = J = J J ev = 1.02 MeV Assuming that the positron has lost its kinetic energy by the time it encounters an electron, by the conservation of energy this will be the combined energy of the two γ photons produced in the annihilation. To conserve momentum, the photons move off in opposite directions, each with equal and opposite energy. This means they must share the 1.02 MeV equally, i.e. 5 kev each a) 82Pb Bi 210 8Po 82Pb β β α [Total 20 marks] 206 [3] b) It was necessary to pump out the air because α-particles are absorbed by a few centimetres of air and would otherwise not even reach the beryllium target. c) As neutrons are particles without any electric charge they cannot cause ionisation and so could not be detected directly by the ionisation chamber. d) In order to conserve both momentum and, as it is an elastic collision, kinetic energy, the neutron must give up all its energy to the proton. The neutron will be brought to rest and the proton will be ejected with the momentum and kinetic energy of the neutron. This is shown in the figure below.

7 e) E γ = ½E p m p c 2 = [0.5 ( ev) ( J ev 1 ) = J = ( kg) ( m s 1 ) 2 ] ½ J J J ev 1 = ev 50 MeV This is an extremely high energy for a photon and very unlikely to be produced as the result of α-particles of energy 5.7 Mev bombarding the beryllium. f) Binding energy is given by: ΔE = c 2 Δm = ( m s 1 ) 2 ( ) kg = J J = J ev 1 = ev This is about 1 to electron volts as Chadwick had predicted. g) In neutron decay: n 0 1 p e ν e The electron anti-neutrino is necessary to conserve lepton numbers and spin. Charge is conserved (= 0) as is the number of baryons (1). [3] h) In terms of quarks, a proton is uud and a neutron is udd so, in neutron decay, an up quark changes into a down quark. This involves the weak force. [Total 20 marks]