Question 38
38
A class of students used dice to simulate radioactive decay. After each throw, those dice showing a ' 6 ' were removed. The graph shows the results. What could the scatter of points about the best-fit curve represent for actual radioactive decay? background count not being taken into account more than one type of radiation being present the random nature of radioactive decay the spontaneous nature of radioactive decay
Easymcq1 marks
Answer
C
Question 7
7
A polonium nucleus \({ }_{84}^{210} \mathrm{Po}\) is radioactive and decays with the emission of an \(\alpha\)-particle. The nuclear reaction for this decay is given by
structured1 marks
Question 7(b)
7(b)
The reaction is spontaneous. Explain the meaning of spontaneous.
Easystructured1 marks
Answer
not affected by external conditions/factors/environment or two examples temperature and pressure B1 [1]
Question 40
40
In a radioactive decay series, three successive decays each result in a particle being emitted. The first decay results in the emission of a \(\beta\)-particle. The second decay results in the emission of an \(\alpha\)-particle. The third decay results in the emission of another \(\beta\)-particle. Nuclides P and S are compared. Which statement is correct? P and S are identical in all respects. P and S are isotopes of the same element. S is a different element of lower atomic number. S is a different element of reduced mass.
Mediummcq1 marks
Answer
B
Question 40
40
A counter recording radioactive decays from a radioactive source gives the following counts in equal intervals of time. What can be deduced from these readings? that radioactivity is random and that the half-life is 90 minutes that radioactivity is random and that the half-life is uncertain that radioactivity is spontaneous and that the half-life is 90 minutes that radioactivity is spontaneous and that the half-life is uncertain
Mediummcq1 marks
Answer
B
Question 8
8
The element strontium has at least 16 isotopes. One of these isotopes is strontium-89. This isotope has a half-life of 52 days.
structured5 marks
Question 8(b)
8(b)
Calculate the probability per second of decay of a nucleus of strontium-89. \(\mathrm{s}^{-1}\)
Mediumstructured3 marks
Answer
probability of decay per unit time is the decay constant C1 \(\lambda=\ln 2 / \mathrm{t}_{1 / 2}\)
Question 8(c)
8(c)
A laboratory prepares a strontium- 89 source. The activity of this source is measured 21 days after preparation of the source and is found to be \(7.4 \times 10^{6} \mathrm{~Bq}\). Determine, for the strontium-89 source at the time that it was prepared,
structured2 marks
Question 8(c)(i)
8(c)(i)
the activity, Bq
Mediumstructured2 marks
Answer
\(A=A_{0} \exp (-\lambda t)\) (alternative method uses 21 days as 0.404 half-lives)
Question 8(c)(ii)
8(c)(ii)
the mass of strontium-89. Answer all the questions in the spaces provided.
Hardstructured0 marks
Answer
\(A=\lambda N\) and mass \(=N \times 89 / N_{\mathrm{A}}\) mass \(=\left(9.8 \times 10^{6} \times 89\right) /\left(1.54 \times 10^{-7} \times 6.02 \times 10^{23}\right)\)
Question 8
8
The power for a space probe is to be supplied by the energy released when plutonium- 236 decays by the emission of \(\alpha\)-particles. The \(\alpha\)-particles, each of energy 5.75 MeV , are captured and their energy is converted into electrical energy with an efficiency of 24 %.
structured6 marks
Question 8(a)
8(a)
Calculate
structured2 marks
Question 8(a)(ii)
8(a)(ii)
the number of \(\alpha\)-particles per second required to generate 1.9 kW of electrical power. number per second = \(\mathrm{s}^{-1}[2]\)
Mediumstructured2 marks
Answer
number \(=1900 /\left(9.2 \times 10^{-13} \times 0.24\right)\)
Question 8(b)
8(b)
Each plutonium-236 nucleus, on disintegration, produces one \(\alpha\)-particle. Plutonium- 236 has a half-life of 2.8 years.
structured2 marks
Question 8(b)(i)
8(b)(i)
Calculate the decay constant, in \(\mathrm{s}^{-1}\), of plutonium-236. decay constant = \(\mathrm{s}^{-1}[2]\)
Mediumstructured2 marks
Answer
decay constant \(=0.693 /(2.8 \times 365 \times 24 \times 3600)\)
Question 8(b)(ii)
8(b)(ii)
Use your answers in (a)(ii) and (b)(i) to determine the mass of plutonium-236 required for the generation of 1.9 kW of electrical power.
Hardstructured0 marks
Answer
\(A=\lambda N\) \(8.6 \times 10^{15}=7.85 \times 10^{-9} \times N\) \(N=1.096 \times 10^{24}\) mass \(=\left(1.096 \times 10^{24} \times 236\right) /\left(6.02 \times 10^{23}\right)\)
Question 8(c)
8(c)
The minimum electrical power required for the space probe is 0.84 kW . Calculate the time, in years, for which the sample of plutonium-236 in (b)(ii) will provide sufficient power. time = years [2] Answer all the questions in the spaces provided.
Hardstructured2 marks
Answer
\(0.84=1.9 \exp \left(-7.85 \times 10^{-9} t\right)\) \(t=1.04 \times 10^{8} \mathrm{~s}\) =3.3 years
Question 9
9
8 marks
Question 9(a)
9(a)
3 marks
Question 9(a)(i)
9(a)(i)
State what is meant by the decay constant of a radioactive isotope.
Easystructured0 marks
Answer
either probability of decay (of a nucleus) M1 per unit time A1 or \(\quad \lambda=(-)(\mathrm{d} N / \mathrm{d} t) / N\) (-) d N / d t and N explained
Question 9(a)(ii)
9(a)(ii)
Show that the decay constant \(\lambda\) and the half-life \(\frac{t_{1}}{2}\) of an isotope are related by the expression
Mediumstructured3 marks
Answer
in time \(t_{\frac{1}{2}}\), number of nuclei changes from \(N_{0}\) to \(\frac{1}{2} N_{0} \quad\) B1 \(\frac{1}{2}=\exp \left(-\lambda t_{1 / 2}\right) \quad\) or \(2=\exp \left(\lambda t_{1 / 2}\right) \quad\) B1 \(\ln (1 / 2)=-\lambda t_{1 / 2}\) and \(\ln (1 / 2)=-0.693 \quad\) or \(\quad \ln 2=\lambda t_{1 / 2}\) and \(\ln 2=0.693 \quad\) B1 \(0.693=\lambda t_{1 / 2}\) A0
Question 9(b)
9(b)
In order to determine the half-life of a sample of a radioactive isotope, a student measures the count rate near to the sample, as illustrated in Fig. 9.1. Initially, the measured count rate is 538 per minute. After a time of 8.0 hours, the measured count rate is 228 per minute. Use these data to estimate the half-life of the isotope. half-life = hours
Mediumstructured3 marks
Answer
\(228=538 \exp (-8 \lambda) \quad \mathrm{C} 1\) \(\lambda=0.107\) ( hours \(^{-1}\) ) C1 \(t_{1 / 2}=6.5\) hours (do not allow 3 or more SF) A1
Question 9(c)
9(c)
The accepted value of the half-life of the isotope in (b) is 5.8 hours. The difference between this value for the half-life and that calculated in (b) cannot be explained by reference to faulty equipment. Suggest two possible reasons for this difference. 1 2. Answer all the questions in the spaces provided.
Mediumstructured2 marks
Answer
e.g. random nature of decay background radiation daughter product is radioactive (any two sensible suggestions, 1 each) B2 Page 6 Mark Scheme:Teachers'version Syllabus Paper GCE AS/A LEVEL-May/June 2012 9702