2.3.9 radioactive decay

Discussion Questions What is radioactive decay? Why does it happen? What is the definition of half life? How much of the original sample is left after two half lives? What is activity? What is the unit of activity? Are count rate and activity the same thing?

Radioactive Decay It’s really all about unstable isotopes changing to stable isotopes. It is random. One element is changing into another...and an α or a β particle is emitted.

Half-Life The half-life, T1/2, of a radioactive isotope is the time taken for the mass of the isotope to decrease to half the initial mass. If the half-life is very long then the substance will remain radioactive for a very long time. If the half life is short it will decay quickly

Decay Curve

Activity The activity, A ,of a radioactive isotope is the number of nuclei of the isotope that disintegrate per second. Activity is measured in Becquerels (Bq) which is the number of decays per second.

Count Rate versus Activity They not quite the same thing. Activity is the number of nuclei that decay per second. Count rate is the number of decays detected per second (and includes background counts).  They love to trap you with this!

Theory of Radioactive Decay Let N0 be the no. of nuclei of a radioactive sample X. Let N be the no. of nuclei of X left after time t. In time Δt, a number of nuclei disintegrate, ΔN Because ΔN is proportional to N and Δt:- Where λ is the constant of proportionality and is known as the decay constant.

The Decay Constant The decay constant is the probability of an individual nucleus decaying per second. Is has the symbol λ, but has nothing to do wavelength (they just share symbols) It has units of s-1 (per second).

Theory of Radioactive Decay Re-arranging the last equation:- And recall that ΔN/Δt is the rate of disintegration which is the activity. So the activity A, of a sample of N nuclei is given by:

Exponential Decay Laws Where N is the number of nuclei of isotope X after time t and N0 is original no. of nuclei. Where m is the mass of an isotope X after time t and m0 is the original mass of isotope X.

Exponential Decay Laws Where A is the activity of the sample after time t and A0 is original activity (t=0) of the sample. Where C is the corrected count rate of the sample after time t and C0 is the original (t=0) corrected count rate of the sample.

Half-Life II The time it takes mass of a radioactive sample to decrease by 50%. The time it takes the activity of a radioactive sample to decrease by 50%. The time it takes the count rate from a radioisotope to decrease by 50%.

Avogadro’s Constant One mole of any gas contains the same number of particles. This number is called Avogadro’s constant and has the symbol NA. The value of NA is 6.02 × 1023 particles per mole.

Calculating the Number of Moles The number of moles, n, of a gas can be can be calculated using:- Where N is the total number of molecules and NA is Avogadro’s constant (=6.02 × 1023)

Calculating the Number of Moles The number of moles can also be calculated from the mass:- Where m is the total number of molecules and MS is the molar mass (the mass of 1 mole of the substance, = nucleon number in grams)

Calculating the Number of Moles You can calculate the number of atoms in the sample from the mass using:- You can also go the other way  calculate the mass from the number of atoms.

Practice Questions 34g of Carbon-14  number of atoms? 0.534 kg of Uranium-235  number of atoms? 234mg of Calcium-40  number of atoms? 5.4 × 1023 atoms of Oxygen-18  mass = ? 3.87 × 1020 atoms of Deuterium  mass = ? 7.84 × 1027 atoms of Aluminium-27 mass=?

Question 1 0.25 kg radon-226 emits alpha particles at a measured rate of 9 × 1012 s-1. What is the decay constant of radium? (No of atoms in a mole = 6 × 1023)

Answer 1 Work out the number of particles: 0.250 × 6 × 1023 = 6.64 × 1023 atoms 0.226 We know that the rate of decay is 9 × 1012 s-1. So we use DN/Dt = -λN - 9 × 1012 s-1 = -λ × 6.64 × 1023 λ = 1.36 × 10-11 s-1 (The minus sign indicates a decay)

Question 2 A sample of living material contains carbon 14 with an activity of 260 Bq kg-1. What is the decay constant? (The fraction that is made of carbon-14 is 1.4 ´ 10-12)

Answer 2 A sample of living material contains carbon 14 with an activity of 260 Bq kg-1. What is the decay constant? (The fraction that is made of carbon-14 is 1.4  10-12) No of particles = 1000/14 × 6 × 1023  1.4  10-12 = 6.02 × 1013 (P) Use DN/Dt = -lN (P) -260 = -l × 6.02 × 1013 l = 4.32 × 10-12 s-1 (P)

Question 3 A radiographer has calculated that a patient is to be injected with 1 x 1018 atoms of iodine 131 to monitor thyroid activity. The half-life is 8 days. Calculate: (a)    the radioactive decay constant (b)   the initial activity (c)    the number of undecayed atoms of iodine 131 after 24 days. (d)   The total activity after 3 days.

Answer 3 (a) We need to use T1/2 = 0.693 λ we need to convert the 8 days into seconds. λ = 0.693 _ = 1.00 x 10-6 s-1 8 x 86400 (b) Use DN = - λ N = 1.00 x 10-6 s-1 x 1 x 1018 Dt = 1 x 1012 Bq

Answer 3 (c) 24 days is 3 half-lives. Therefore the number atoms remaining undecayed is 1/8 of the original. N = 1 x 1018 / 8 = 1.25 x 1017 (d) 3 is not so easy. We use A = A0e-λt A = 1 x 1012 Bq x e-(1.00  10-6 s-1  3  86400s)   A = 1 x 1012 Bq x e-(0.2592) = 1 x 1012 Bq x 0.772 = 7.72 x 1011 Bq.

Question 4 What is the half life of radon-226? λ = 1.36 × 10-11 s-1

What is the half life of radon-226? l = 1.36 × 10-11 s-1   Formula: N = N0e-lt ½ = e-1.36  10^-11  t1/2 (P)   Loge ½ = -1.36  10-11  t1/2 (P) t1/2 = -0.693  -1.36  10-11 = 5.1  1011 s (= 1600 years) (P) Note 10^11 means 1011.  I cannot do a double superscript. 

Question 5 Strontium-90 is a beta emitter Question 5 Strontium-90 is a beta emitter. It is one of the radio-nuclides found in the fall out from an atomic bomb explosion. It can be absorbed into the bone. It emits beta particles and has a half life of 28 years. What is the time needed for the activity to fall to 5 % of the original?

Strontium-90 is a beta emitter Strontium-90 is a beta emitter. It is one of the radio-nuclides found in the fall out from an atomic bomb explosion. It can be absorbed into the bone. It emits beta particles and has a half life of 28 years. What is the time needed for the activity to fall to 5 % of the original? We need to know how many half lives gives 5 % (0.5)y = 0.05 (P) y  log (0.5) = log (0.05)

y  -0.3010 = - 1.3010 y = 1.3010  0.3010 = 4.32 half lives (P) Time taken = 4.32  28 (P) Time taken = 121 years (P)

Question 6 A GM tube placed close to a radium source gives an initial average corrected count rate of 334 s-1 (a)     The GM tube detects 10 % of the radiation. What is the initial activity? (b)    Initially there were 1.5 ´ 109 nuclei in the sample. What is the decay constant? (c)     What is the half life of the radium in days?

A GM tube placed close to a radium source gives an initial average corrected count rate of 334 s-1 (a)     The GM tube detects 10 % of the radiation. What is the initial activity? (b)    Initially there were 1.5  109 nuclei in the sample. What is the decay constant? (c)     What is the half life of the radium in days? (a) The activity is 334  0.1 = 3340 s-1 (P)

(b) DN/Dt = -lN (P) -3340 = -l  1.5  109 = 2.23  10-6 s-1 (P) (c) N = No e-lt ½ = e-2.23  10^-6 t loge ½ = -2.23  10-6  t (P) t = -0.693  2.23  10-6 = 3.1  105 s t = 3.6 days (P)

Summary Radioactive decays are random. Rate of decay depends on the number of atoms left The probability of any one nucleus decaying in any one second is the decay constant Decay constant is given the code l DN = -lN Dt

Over a longer period of time, decay is exponential. N = No e-lt Half life is the time taken for ½ the remaining atoms to decay T1/2 = 0.693