Radioactive Decay PDF

2 Radioactive Decay By: Abdullah Munqith Decay process(Activity) Radioactive decay is a random process. We cannot predict when an individual nucleus will decay but with large numbers of nuclei we can use a statistical approach. One of the most important quantities associated with a sample of radioac...

2 Radioactive Decay By: Abdullah Munqith Decay process(Activity) Radioactive decay is a random process. We cannot predict when an individual nucleus will decay but with large numbers of nuclei we can use a statistical approach. One of the most important quantities associated with a sample of radioactive material is its activity. Activity is the rate at which the nuclei within the sample undergo decay (disintegrations) and can be expressed in terms of the number of disintegrations per second (dps). Activity Activity : It is the average number of disintegrations per second (dps). The SI unit of activity is the becquerel: 1 becquerel (1 Bq ) = 1 disintegration/second ( 1dis/sec). 1 Bq= 1dis/sec= 1dps Another unit of activity is the curie (Ci): 1 curie = 1 Ci = 3.7x1010 dis/s 1 Ci = 3.7x1010 Bq Radioactive Decay Law Experimental measurements show that the activities of radioactive samples fall off exponentially with time. As the sample decays, less of the radioactive sample remains, so the activity decreases. Radioactive Decay Law The following equations can be used to calculate the activity of a radioactive material at any time. A = A0 e-λt A = λN A0 = λN0 A = Activity remaining in the radioactive material after time(t). A0 = Initial Activity of the radioactive material. N = Number of radioactive nuclei present at a time t. N0 = the initial number of radioactive nuclei present at time t=0. λ = Decay constant. t = Time. e = Euler's number Radioactive Decay Law We can also calculate the number of remaining radioactive nuclei in the radioactive sample after period of time from the following equation: Nt = N0e-λt N = Number of radioactive nuclei present at a time t. N0 = the initial number of radioactive nuclei present at time t=0. λ = Decay constant (s-1). t=Time (s). Radioactive Decay Law Decay constant (λ): The probability that a nucleus will decay per unit of time. Its unit is s-1 , min-1 , h-1 , day-1 s-1 = 1 s Half-Life (t1/2): The time for half of the radioactive nuclei in a given sample to undergo decay. (i.e. the time it takes for the activity to drop by 1⁄2) Radioactive Decay Law Each radioactive nuclide has a particular decay constant and half life. The relationship between λ and t1⁄2 is: λ = ln 2 t1/2 λ = 0.693 t1/2 Radioactive Decay Law The initial activity was chosen to be 1000 for this plot. The half-life is 10 (in whatever time units we are using). Note: All decay curves look like this; only the numbers on the axes will differ, depending on the radionuclide (which determines the half-life) and the amount of radioactive material (which determines the initial activity). a plot of the activity of a radionuclides versus time. Radioactive Decay Law a plot of number of a radionuclides versus half life. Common Radioactive Isotop Isotope Half-Life Radiation Emitted Technetium-99 6 hours 𝛾 Radon-222 3.8 days 𝛼 Carbon-14 5,730 years 𝛽, 𝛾 Uranium-235 7.0 x 108 years 𝛼, 𝛾 Uranium-238 4.46 x 109 years 𝛼 Note: You do not need to memorize these number Examples Ex.1: A sample of 3x107 Radon atoms are trapped in a basement that is sealed. The half-life of Radon is 3.83 days. How many radon atoms are left after 31 days? Answer: Info: N0 = 3x107 atom, t1⁄2 = 3.83 days, t=31 days To find N after 31 days, we will use this eq. Nt = N0e-λt λ = (0.693/3.83 d) = 0.1809 d-1 Next step, we will find N by using: Nt = N0e-λt Nt = 3x107 atom x exp(- 0.1809 d-1 x 31d) Nt =1.1x105 atom First, we should calculate λ from this eq: λ = 0.693 t1/2 Examples Ex.2: We received 10 millicuries (mCi) of I-125. The half-life of I-125 is 60 days. How much activity remains after 3 months. Answer: Info: A0 = 10 m Ci, t1⁄2 = 60 d, t = 3 months= 90 d we are asked to calculate At= ? Therefore, we will use this eq. At = A0 e-λt t1⁄2 = 60 d λ = (0.693/60 d ) = 0.01155 d-1 At = A0 e-λt At =10mCi x exp(-0.01155x90) At =3.53 mCi the activity after 3 months Examples Ex.3: A sample of 3x106 Radon atoms are trapped in a basement that is sealed. The half-life of Radon is 3.83 days. How much the current activity? Answer: Info: N= 3x106 , t1⁄2 = 3.83 d λ = (0.693/ t1⁄2 ) We are asked to find the activity (A) ?? Therefore, we will use this eq: A= λN Before solving this problem, you have to make the unit of λ in s-1 t1⁄2 = 3.83 d x 24 h/d x 60 min/h x 60 s/min = 330912 sec λ = (0.693/330912) = 2 x10-6 s-1 A= 2 x10-6 s-1 x 3 x106 atom = 6 atom/s = 6 Bq Examples Ex.4: A sample of Radon has an activity of 10 Bq. The half-life of Radon is 3.83 days. How many radioactive nuclides are in this sample? Answer: Info: A =10Bq=10dis/sec, t1⁄2 =3.83d A= λN t1⁄2 = 3.83 d x 24 h/d x 60 min/h x 60 s/min = 330912 sec λ = (0.693/330912) = 2 x10-6 s-1 N= A/λ = (10dis/sec)/(2 x10-6 s-1) N = 5 x106 atoms Examples Ex.5: A medical center received Tc-99 (half life = 6 h). The initial activity of Tc-99 was 3 mCi when it was delivered,. Calculate the activity of Tc-99 after 2 days ? Answer: Info: A0 =3mCi, t1⁄2 =6h, t=2d=48h At = A0 e-λt t1⁄2 = 6 h λ = (0.693/6 h ) = 0.1155 h-1 At = 3 mCi x exp( - 0.1155 x 48) = 3mCi x 0.0039 At = 0.0117 mCi the activity after 2 days Radiation Dose ▪ The magnitude of radiation exposures is specified in terms of the radiation dose. ▪ There are two important categories of dose: 1. 2. Absorbed dose. Dose equivalent. Radiation Dose 1- Absorbed dose Absorbed dose: the amount of energy deposited in a unit mass in human tissue or other media. SI unit used to measure absorbed dose is the gray (Gy) 1 Gy = 1 J/kg Another unit used to measure absorbed dose is rad 1 Gy = 100 rad Absorbed dose can be calculated from the following formula: D = E/m Where: D=dose (Gy), E =energy(J), m= mass(Kg) Radiation Dose Example: 0.1 J of radiation energy was deposited in 30g of tissue. Find the absorbed dose in Gy and in rad. Answer Info: E=0.1 J, m=30g =30/1000 =0.03 Kg D = E(J)/m(Kg) D = 0.1J/0.03Kg = 3.3 Gy Since 1 Gy = 100 rad D = 3.3 Gy x 100 rad/Gy D= 330 rad Radiation Dose 2- Dose equivalent Dose equivalent: A measure of the biological damage to living tissue as a result of radiation exposure. Also known as the " biological dose”. SI unit used to measure Dose equivalent is sievert (Sv) Another unit used to measure Dose equivalent is rem 1 Sv = 100 rem Dose equivalent can be calculated from the following formula: H (Sv) = D (Gy) x Q H (rem) = D (rad) x Q Radiation Dose Where: H = Dose equivalent , D= absorbed dose Q= weighting factors(previously called quality factor) Weighting factor (Q): is dimensionless factor used to convert physical dose (Gy) to equivalent dose (Sv). Radiation Dose Example: If the absorbed dose is 20 rad and Q is 10. What is Dose equivalent in rem and in Sv Answer: H (rem) = D( rad) x Q = 20 rad x 10 = 200 rem Since 1Sv=100rem →1rem=(1/100)Sv=0.01Sv H (Sv) = 200 rem x 0.01 Sv/rem= 2 Sv Or, we can solve it by this: H (Sv) = D (Gy) xQ =20 rad x(1/100) Gy/rad x10 H = 0.2 Gy x 10 = 2 Sv Factors Affecting Radiation Dose Target organ/tissue ,i.e.,radiation sensitivity of the tissue Type and energy of the radiation Rate of Delivery Interaction Volume Biological status: difference between young/old, male /female ,population/individual, healthy/diseased Radiation Interaction In Cell There are two principle ways to characterize the ways in which radiation interacts with cells. These ways are through direct effects and indirect effects. 1- Direct Effect -Direct ionization of DNA or other structure by radiation. The result may lead to the break-up or chemical change in the molecule. 2-Indirect Effect- Ionization of water molecules within the cell creates very chemically active agents (free radicals) which can chemically attack other molecules in their immediate vicinity. If a DNA molecule is located in this area, alteration to the DNA molecule could occur. Radiation Safety and ALARA What is ALARA? ALARA is an acronym for “As Low As Reasonably Achievable”. This is a radiation safety principle for minimizing radiation doses and releases of radioactive materials by employing all reasonable methods. ALARA is not only a sound safety principle, but is a regulatory requirement for all radiation safety programs. Three of the most basic and easy to follow principles of radiation protection are time, distance, and shielding. We can greatly reduce our exposure by following these principles. Radiation Safety and ALARA Time: As the length of time a person is exposed increases, the dose received increases. Distance: The most effective of the principles is distance. The further a person is from the source the less intense the radiation source is. Shielding: When the use of the time and distance principles are not possible shielding should always be used. Wearing protective lead shielding and thyroid collars can protect the radiosensitive areas of the body when it is required for the technologist to be near the source of radiation. Reducing External Radiation Exposure Time: reduce time spent in radiation area. Distance: stay as far away from the radiation source as possible. Shielding: interpose appropriate materials between the source and the body Inverse square law applied in radiation Inverse Square law: The radiation Intensity is inversely proportional to the square of the distance from the source. The radiation source must be a point source I1 xD1 2 =I2 xD22 I1 = Intensity at a distance D1 measured as (R/hr or mR/hr). D1 = first Distance. I2 = Intensity at Distance D2. D2 = second Distance. Inverse square law applied in radiation Inverse square law applied in radiation Inverse square law applied in radiation Example 1: If the intensity of point source = 100 IU at 1cm, what is its intensity at 10 cm? Answer: Given Information: I1=100 IU, D1=1 cm, D2=10 cm, I2=? I1 xD1 2 =I2 xD22 100 x (1)2 = I2 x (10)2 100 = I2 x 100 I2=1 IU IU= Intensity Unit Inverse square law applied in radiation Example 2: A reading of 100 mR/hr is obtained at a distance of 1 cm from a point source. What would be the reading at a distance of 1 mm? Answer: Given information: I1= 100 mR/hr, D1= 1 cm, D2=1 mm =0.1 cm, I2=? I1 xD1 2 =I2 xD22 100 x (1)2 = I2 x (0.1)2 100 = I2 x 0.01 I2 = 100/0.01 I2 = 10000 mR/hr