Practical Examples Related To Above Topics
| Let us discuss various examples in order to get a complete knowledge on the above topics. | ||||||||
| Ex.1 Let us see the application of basic radioactive law here and the quantities associated with this. Q. If any radioactive sample has half life 50 days, what is the disintegration constant and average life? How much time is required when 1/4th of the original number of atoms remain unchanged? | ||||||||
| Ans. We know that disintegration constant λ = 0.693/half life (T1/2). Here T1/2 = 50 days. Therefore λ = 0.01386 day-1. Mean or average life Tmean = 1/λ ≅ 72 days. Now if 1/4th of the original number of atoms have to remain unchanged N/N0 = ¼. Now from the basic equation:
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| Ex.2 In this we will draw the decay curves for parent and first daughter comparing their decay constants. Q. Plot activity of parent (AP) and daughter (AD) in case of following radioactive decay: P → D → Stable considering following cases: (a) short-lived parent (λP >λD), (b) long-lived parent (λP <λD), (c) very long lived parent (λP << λ D), (d) almost stable parent (λ P ≅ 0). | ||||||||
Ans.
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| Ex.3 In this we will try to understand how to determine the weight of any element that is in secular equilibrium with parent of any radioactive series. Q. Calculate the amount of Ra-226 in secular equilibrium with 5 kg of pure U-238, given the half life of 226Ra and 238U as 1620 years and 4.5 x 109 years respectively. | ||||||||
| Ans. | ||||||||
| Let x gm be the required amount of 226Ra in secular equilibrium with 5 kg of pure 238U. Now, | ||||||||
5 kg of pure U-238 = /equ18.png){kind=small} atoms of U | ||||||||
x gm of Ra-226 = /equ19.png){kind=small} atoms of Ra | ||||||||
| From the condition of secular equilibrium | ||||||||
/n1.png) | ||||||||
| x = 1.7 mg | ||||||||
| Ex.4 this example we will see how AMS separates C-14 from its non-radioactive counterparts. Q. An AMS running with a terminal potential of 10 MeV accelerates C-12, C-13 and C-14 (completely stripped off all electrons) towards analyzer stage | ||||||||
| Ans. The charge state of all isotopes is 6 and therefore energy [(q+1) x T.P.] gained by all isotopes is also equal (70 MeV). After passing through analyzer magnet each ion will be bent into a track of radius. R = (mv/Bq)1/2 where B is the analyzer field strength. Again, 1/2 mv2 = 70 MeV v = (140/m)1/2 Therefore, R α m1/4 and the isotope with highest mass will make a deviated track of highest radius. | ||||||||
| Ex.5 In this example we will understand the radiation dose absorbed by any person working with a radioactive source. Q. A radiobiologist working with a 22Na source of strength 100µCi keeping it at a distance of 60 cm. Suppose skin absorbs 90 erg/g for 1R γ radiation, what is the effective dose rate received by his skin? If he works continuously for 5 hours, comment whether he has done his job safely. Exposure rate of any radiation source is defined as : Q.A/d2, where Q is the exposure rate constant, A is the activity of the source and d is the distance.. For Na-22, Q = 12 [R-cm2/hr-mCi]. | ||||||||
| Ans. In the present case Exposure rate = 12 x 0.1/602 = 0.33 mR/hr. Dose rate = 90 x 0.33 x 10-3 = 0.030 erg/g-hr = 0.030 mGy/hr. For 5 hours of exposure absorbed dose = 0.15 mGy. Normalizing with radiation and tissue weighting factor, the effective dose becomes 0.15 x 1 x 0.01 = 0.0015 Sivert. This dose is much below the threshold dose that can affect skin. | ||||||||
| Ex.6 In this example we will study some of the basic aspects of NMR and MRI . Q. A biological sample containing 0.1 cc of water is kept in a MRS scanner. If 0.1 Tesla magnetic field and a pulsed (pulse rate 1 will be generated in the second) RF (perpendicular to magnetic field is used) , what is maximum power generated in the output circuit when 100% spin flipping takes place. | ||||||||
| Ans: Larmor precessional frequency of proton under this magnetic field is:
Energy absorbed or released when a proton flips from one orientation to ther, E = h νLp = 6 x 10-34 x 4 x 106 Joules = 2.4 x 10-27 Joules = 1.5 x 10-8 eV. Now number of protons in 0.1 cc of water is 1022, and hence for an ideal case when all protons are flipped the energy involvement is 1.5 x 10-8 eV x 1022 = 1.5 x 1014 eV = 2.4 x 10-5 Joules. As the RF pulse rate is 1 sec the power involved in this process is 2.4 x 10-5 watt or 24 milliwatt. |
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cycles/sec, where g value of proton is 5.56.