Question
Polonium-\(211\) \(\left ( _{84}^{211}\textrm{Po} \right )\) decays by alpha emission to form a stable isotope of lead (Pb).
(a) Complete the equation for this decay.
(b) The variation with time t of the number of unstable nuclei N in a sample of polonium-\(211\) is shown in Fig. 9.1.
At time t = 0, the sample contains only polonium-\(211\).
(i) Use Fig. 9.1 to determine the decay constant λ of polonium-\(211\). Give a unit with your answer.
(ii) Use your answer in (b)(i) to calculate the activity at time t = 0 of the sample of polonium-\(211\).
(iii) On Fig. 9.1, sketch a line to show the variation with t of the number of lead nuclei in the sample.
(c) Each decay releases an alpha particle with energy \(6900\) keV.
(i) Calculate, in J, the total amount of energy given to alpha particles that are emitted between time \(t = 0.30\) s and time \(t = 0.90\)s.
(ii) Suggest why the total amount of energy released by the decay process between time t = \(0.30\) s and time t = \(0.90\) s is greater than your answer in (c)(i).
Answer/Explanation
Ans:
(a) \(207\), \(82\) for lead
\(4\), \(2\) for alpha
(b) (i) (half-life found as) \(0.52\) s or correctly read points substituted into
\(N\) = \(N_{0}e^{-2t}\)
\(\lambda\) = \(\frac{0.693}{t_{1/2}}\)
\(\lambda\) = \(\frac{0.693}{0.52}\)
\(\lambda\) = \(1.3\, s^{-1}\)
(ii) \(A\) = λN
\(12\) = \(1.3 \times 24 \times 10^{12}\)
\(13\) = \(3.1 \times 10^{13}\) Bq
(iii) upwards curve of decreasing gradient starting from \((0,0)\)
passes through \((0.52, 12)\) and \((1.2, 18.8)\)
(c) (i) \(16 \times 10^{12}\) and \(7.2 \times 10^{12}\)
\(6900 \times 10^{3} \times 1.6 \times 10^{-19}\)
\((16 \times 10^{12} – 7.2 \times 10^{12}) \times 6900 \times 10^{3} \times 1.6 \times 10^{-19}\)
= \(9.7\) J
(ii) lead nuclei have kinetic energy
or
gamma photons
Question
(a) Define radioactive decay constant. [2]
(b) A sample of radioactive iodine-131 of mass 5.87 × 10–10 kg has an activity of 2.92 × 109Bq.
Determine the decay constant of iodine-131.
decay constant = …………………………………………… s–1 [3]
(c) Suggest two reasons why a detector placed near to the sample in (b) would record a count rate much less than 2.92 × 109 counts per second.
1.
2.
[2]
[Total: 7]
Answer/Explanation
Ans
(a) probability of decay (of a nucleus)
per unit time
(b) A = λN
N = mass / (nucleon number × u)
2.92 × 109 = ( λ × 5.87 × 10–10) / (131 × 1.66 × 10–27)
λ = 1.08 × 10–6 s–1
(c) • sample emits radiation in all directions
• some radiation is absorbed by air/detector window
• self-absorption within the source
• dead time/inefficiency of detector