Practical Lecture PDF

Practical Lecture Dr. Nahla Atallah Question: What is the maximum number of electrons that can exist in the O shell? Answer: The O shell is the fifth shell from the nucleus; therefore: ⸪n=5 ⸫ 2n2 = 2(5)2 = 2(25) = 50 electrons..  Units...

Practical Lecture Dr. Nahla Atallah Question: What is the maximum number of electrons that can exist in the O shell? Answer: The O shell is the fifth shell from the nucleus; therefore: ⸪n=5 ⸫ 2n2 = 2(5)2 = 2(25) = 50 electrons..  Units of radioactivity. Curie 1 Ci=3.7x1010 disintegration/sec. Bequral Bq= 1 disintegration/sec. 1 Ci = 3.7x1010 Bq. Que: Which element of the following is β- or β+ emitter: 20Ca42, 11Na21? Answer: let: N = number of neutrons. Z = number of protons. For 20Ca42, N = 42 – 20 = 22 but Z = 20 So N > Z then 20Ca42 is β- emitter. For 11Na21, N=21-11=10 but Z=11 So N< Z then 11Na21 is β+ emitter  Half life time T1/2 The time needed for all of the radioactive nuclei to decay to its half activity. A=Aoe- λt -------------------- (1) At t = T1/2 ----------- A=1/2 Ao subst. in eqn. (1) ------ ½ Ao=Ao e- λT1/2 2= e λT1/2 ln 2 = λ T1/2 T1/2 = 0.693/λ Effective Half-Life Te It is defined as the time needed for half of the radiopharmaceutical to disappear from the biologic system. Question: If a piece of petrified wood contains 25% of the 14C that a tree living today contains, how old is the petrified wood? Answer: The 14C in living matter remains constant as long as the matter is alive because it is constantly exchanged with the environment. In this case, the petrified wood has been dead long enough for the 14C to decay to 25% of its original value. That time period represents two half-lives. Consequently, we can estimate that the petrified wood sample is approximately 2 × 5730 = 11,460 years old. Problems 1. If a radionuclide decays at rate of 30 % / h. what is its half-life? Answer: λ = 0.3 hr -1 0.693  t1 / 2 0.693 0.693 t1/ 2   hr  2.31hr  0.3 2. If 11 % of 99mTc-labeled diisopropyliminodiacetic acid (DISIDA) is eliminated via renal excretion. 35% by fecal excretion, and 3.5% by perspiration in 5 hr from the human body, What is the effective half-life of the radio- pharmaceutical (Tp = 6 hr for 99mTc) ? Answer: Total biological elimination = 11% + 35% + 3.5% = 49.5 % in 5 hr » Therefore, Tb ≈ 5 hr Tp = 6 hr Tb  Tp 5  6 30 Te     2.7hr Tb  Tp 5  6 11 » 3. Calculate the time required to reduce the activity of a pure 40K from 7 µ Ci to 50 kBq? (T1/2=2days)? Given: T1/2=2 days Ao=7µCi=7x10-6x3.7x1010Bq A=50kBq=50x103Bq Required: t ????? Equations and calculations: T1/2=0.693/λ λ = 0.693/2 = 0.3465 day-1 A=Ao e-λt 50x103 = 7x3.7x104 e-0.3465t 50=259 e-0.3465t t= (ln 259/50)/0.3465 = 4.75 days.