Umm Al-Q ura University جا معة أم القرى
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1- Source activity
3- Absorbed dose
4- Biologically equivalent dose
1 curie= 1 Ci = 3.7 x 1010 disintegration per second
A = - N/ t = N = 0.693 N/T
where Avogadro’s number NA = 6.02 x 1023 is the number of particles in a mole.
A = 0.693 n NA/T
60Co beta decays with a half-life of 5.27 years = 1.66x108s into 60Ni, which then promptly emits two gamma rays. These gamma rays are widely used in treating cancer.
What is the mass of a 1000 Ci cobalt source?
From the following equation:
A = 0.693 n NA/T
the number of moles n is : n = A T /0.693 NA
n = 1000 (3.7x1010 s-1)(1.66x108 s)/0.693(6.02 x 1o23 mole-1)
Since a mole of 60Co has a mass of 60 g, the mass of the sample is:
m =(0.0147 mole)(60 g mole-1)=0.882 g
1 rad = 0.01 joule per kilogram
Table 31.1: Typical QF values. By definition, the QF is exactly 1 for 200 keV X rays.
Table 31.2: Radiation units
A cancer is irradiated with 1000 rads of gamma rays,
Which have a QF of 0.7.
a- The exposure in roentgen
b-The biologically equivalent dose in rems
a- For gamma rays, the soft tissue dose in rads is approximately equal to the exposure in roentgens, so the exposure is approximately 1000 R.
b- The biologically equivalent dose is the product of the QF and the dose in rads:
o.7 x 1000 rads = 700 rems
A laboratory experiment in a physics class uses a 10 microcurie source. Each decay emits a 0.66 MeV gamma rays.
a- How many decays occur per hour?
b- A 60 kg student standing nearby absorbs 10 percent of gamma rays. What is her absorbed dose in rads in 1 hour?
c- The QF is 0.8. Find her biologically equivalent dose in rems.
a- Since 1 curie products 3.7x1010 decays per second, the number of decays in an hour for a 10 microcurie source is:
N= (10x10-6)(3.7x1010 s-1)(3600s)=1.33x109
b- Each decay produces 0.66 MeV, of which 10 percent is absorbed. Thus the total energy absorbed is:
E=(0.1)(1.33x109)(0.66 MeV)(1.6x10-13 J MeV-1)
= 1.41x10-5 J
Dividing by her mass, the absorbed dose in rads is:
E/m = (1.41x10-5 J)(1rad)/(60 kg)(0.01 J kg-1)
c- Multiplying the dose in rads by the quality factor gives the biologically equivalent dose in rems:
(E/m)(QF)= (2.34x10-5 rad)(0.8)=1.87x10-5 rem
-When radiation passes through living cells, it can alter or damage the structure of important molecules.
- This may lead to the malfunctioning or death of the cells and ultimately to the death of the organism.
-The limited knowledge of the immediate effects on humans of large radiation doses comes from studies of victims of atomic bomb explosions and of occasional accidents.
- For :
*whole body doses < 25 rems No observable effect
*doses >100 rems Damage to the blood-forming tissues becomes evident
*doses >800 rems Severe gastrointestinal disorders
- Studies of low-level doses are difficult and inconclusive, because of the much larger incidence of cancer from other causes.
- Most genetics mutations are harmful.
* some chemicals
* ionizing radiation