Nuclear pharmacy Lecture 3

Radioactive decay cont.

4. Electron capture:  This is an alternative to β+ decay, when the nucleus has a smaller N/Z ratio compared to the stable nucleus (neutron deficient nuclei).  It involves the conversion of a proton into neutron emitting a neutrino.  This process is occurred by capture of an electron from the extranuclear electron shell.  The atomic number of the parent is reduced by 1 in this process. Some examples: 6731Ga + e- → 6730Zn + v 5727Co + e- → 5726Fe + v

 Usually the K-shell electrons are captured because of their proximity to the nucleus, the process is then called K-capture.  This capture causes a deficiency in the K-shell, which is made good by the migration of electron from one of the outer shells. Since the migrant negatron loses energy in the process, the excess energy is emitted in the form of x-ray of an energy characteristic of the product atom (daughter nucleus). Example: 5727Co + e- → 5726Fe + v  As with positron emission the mass number remains unchanged but the atomic number is reduced by one unit.  The probability of electron capture increases with increasing atomic number, because electron shells in these nuclei come closer to nucleus.

Electron Capture Decay with gamma ray emission

I-123 Decay

5. Nuclear fission:  Large nuclei tend to be unstable and split into 2 fragments (occur spontaneously, slow).  For natural uranium the process has a half-life of about 0.9 × 106 years.  If large nuclei are irradiated with neutrons, the resultant neutron capture results in instability and fission occurs immediately.  It is the basis of the operation of the nuclear reactor, in which bulk of (235U or 239Pu) is brought together so that the neutrons emitted in the spontaneous fission captured by other nuclei (self-sustaining chain reaction).

Nuclear Fission

Demonstration of nuclear fission In the first frame, a neutron is about to be captured by the nucleus of a U-235 atom. In the second frame, the neutron has been absorbed and briefly turned the nucleus into a highly excited U-236 atom. In the third frame, the U-236 atom has fashioned, resulting in two fission fragments (Ba-141 and Kr-92) and three neutrons.

6. Isomeric transition: It is a process by which a nuclide decays to isomeric nuclide of lower quantum energy. This decay process occurs in an atom where an atom is in the excited meta state (isomeric state) and decay to the ground state. The extra energy in the nucleus is released by the emission of a gamma ray, returning the nucleus to the ground state. When isomeric states are long lived, they are often referred to as meta-stable state is donated by "m", as in 99mTc. 6 hrs 2.1× 105 yr Nuclear isomers: Nuclides with the same mass and atomic number but different half-lives.

decay scheme of Tc-99m

Nuclear stability (neutron-proton ratios)  For smaller nuclei (Z  20) stable nuclei have a neutron-to-proton ratio close to 1:1.

Neutron-proton ratios  As nuclei get larger, it takes a greater number of neutrons to stabilize the nucleus.  The shaded region in the figure shows what nuclides would be stable, the so-called belt of stability.

Nuclei above this belt have too many neutrons Nuclei above this belt have too many neutrons. They tend to decay by emitting beta particles. Nuclei below the belt have too many protons. They tend to become more stable by positron emission or electron capture.

Nuclear stability

Summary of the rules that are useful in predicting the nuclear stability:  All nuclides with 83 or more protons are unstable with respect to radioactive decay. When the neutron to proton ratio is too large or too small the nucleus is unstable. This ratio is close to 1 for atoms of elements with low atomic number and increases as the atomic number increase. Elements have atomic number 83 with 209 nucleons do not exist as stable isotopes thus polonium with 84 protons, its nuclides are unstable.  Nuclides with even number of protons and neutrons are more stable compared to others. Example: 168O and 178O 168O contains 8 protons and 8 neutrons (more stable and less radioactive). 178O contains 8 protons and 9 neutrons (less stable and more radioactive)

 Nuclei that contain a magic number of proton and neutron seems to be more stable. It is noted that nuclei with certain numbers of protons or neutrons appear to be very stable. These numbers are called the magic numbers and associated with specially stable nuclei. According to this theory, a magic number is the number of nuclear particles in a completed shell of protons and neutrons. For protons the magic numbers are 2, 8,20, 28, 50, 82 . Neutrons have these same magic numbers , as well as the magic number 126.

Kinetics of radioactive decay Radioactive decay equations Decay rate: - Is the time rate at which atoms undergo radioactive disintegration. - Radionuclides are unstable and decay by particle emission, electron capture or gamma ray emission. - The decay of radionuclides is a random process. i.e. one cannot tell which atom from a group of atoms will decay at a specific time. - The average number of radionuclides disintegrating during a period of time. The number of disintegrations/unit time = disintegration rate. = -dN/dt -dN: The change in the number of atoms, N. dt : The change in the time, t.

"e" is the base of natural logarithm = 2.71828  Radioactive decay is a first order process - dN/dt of radionuclide at any time is proportional to the total number of radionuclides present at thet time. - dN/dt (D) = λN where N is the number of radionuclides and λ is a decay constant that is defined as the probability of disintegration per unit time for a single radionuclide. - dN/dt (D) radioactivity or simply the activity of a radionuclide. Rearrange: Where N0 and Nt are the number of radionuclides present at t = 0 and time t, respectively. "e" is the base of natural logarithm = 2.71828

 If we remember the basic equation relating activity to number of nuclei in a sample, A=N, then we can write Plot of radioactivity versus time on a linear graph. The time is plotted in units of half-life. Plot of the data in the previous figure on a semi logarithmic graph, showing a straight-line relationship.

Units of radioactivity: From the knowledge of the decay constant and radioactivity of a radionuclide, D=λN we can calculate the total number of atoms or the total mass of radionuclide present using Avogadro’s number, 1gram-atom = 6.02 × 1023 atoms. Units of radioactivity: Radioactivity is expressed in units called curies. 1 curie (Ci) = 3.7 × 1010 disintegration per second (dps) 1 millicurie (mCi) = 3.7 × 107 disintegration per second (dps) 1 microcurie (μCi) = 3.7 × 104 dps The other unit for radioactivity is becquerel (Bq) which is defined as one disintegration per second. Thus 1 becquerel (Bq) = 1 dps = 2.7 ×10-11 Ci 1 megabecquerel (MBq) = 106 dps = 2.7 × 10-5 Ci Similarly, 1mCi = 3.7× 107 Bq = 37 MBq

Half-life and mean life: Every radionuclide is characterized by a half-life, which is defined as the time required to reduce its initial disintegration rate or activity to one-half. It is usually donated by t1/2 and is unique for a given radionuclide. The decay constant λ of a radionuclide is related to half-life by Another relevant quantity of a radionuclide is its mean life, which is the average life of a group of the radionuclides. It is donated by τ and related to decay constant λ and half-life t1/2 as follows: In one mean life, the activity of radionuclide is reduced to 37% of the initial value.

The physical half-life of 131I is 8. 0 days. A The physical half-life of 131I is 8.0 days. A. A sample of 131I has a mass of 100 μg. How many 131I atoms are present in the sample? Number of atoms N = 4.6 × 1017atoms B. How many 131 I atoms remain after 20 days have elapsed? Nt= N0e−(λt) = (4.6 × 1017 atoms)e−(0.693/8 d)(20 d) = 8.1 × 1016 atoms C. What is the activity of the sample after 20 days? A or D = λN = (0.693/8.0 d)(1/86400 s/d)(8.1 × 1016 atoms) = 8.2 × 1010 atoms/sec = 8.2 × 104 MBq

D. What activity should be ordered at 8 AM Monday to provide an activity of 8.2 × 104 MBq at 8 AM on the following Friday? Elapsed time = 4 days At = A0e−λt 8.2 × 104 MBq = A0e−(0.693/8d)(4d) 8.2 × 104 MBq = A0(0.7072) A0 = 11.6 × 104 MBq must be ordered

131I = 6.02 × 1023 atoms of 131I (Avogadro’s number), Problem: Calculate the total number of atoms and total mass of 131I present in 5 mCi (185 MBq) 131I (t1/2 = 8 days). Answer: λ for 131I = D = 5 × 3.7 × 107= 1.85 × 108 dps Using the equation Since 1 g. atom 131I = 131 g 131I = 6.02 × 1023 atoms of 131I (Avogadro’s number), Mass of 131I in 5 mCi (185 MBq) = = 40.3 × 10-9 g = 40.3 ng Therefore, 5 mCi 131I contains 1.85 × 1014 atoms and 40.3 ng 131I.

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