Obtain the amount of ^{60}_{27} Co necessary to provide a radioactive source of 8.0 m Ci strength. The half-life of ^{60}_{27} Co is 5.3 years.

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Law of Radioactive Decay, Decay Rate, Half-Mean Life, Q&A
In this topic, we will learn about the Laws of Radioactive Decay. ... Now, integrating both the sides of the above equation, we get, ... This rate gives us the number of nuclei decaying per unit time. ... Next, let's find the relation between the mean life τ and the disintegration constant λ. ... Download NCERT Notes and Solutions.
For more information, see Law of Radioactive Decay, Decay Rate, Half-Mean Life, Q&A
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Weak Interaction
to find their form for a single particle moving uniformly in the absence of any external ... j f t) ,. (1 .1.1 ) where x and t are the position and time co-ordinates of points on the wave , ... 1.5 .8 x nd. 0. (. ) We now introduce the variables (27). X = 8 712 mW/h2 and. (1 .5 .9 ) ... Mathematical Physics, C .U .P ., 1956, pp . 325-327 . 1 .7.
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Radiation Safety Manual
Table 1.2 Γ- factor, half-life, photopeak, and half value layer for selected gamma emitters ... For example, the dose rate at 10 cm from a 2 mCi source of Co-57 is:.
For more information, see Radiation Safety Manual
Required formulas:
Decay rate R = \lambda N where, N - number of radioactive atoms and \lambda - decay constant
Half life of the radioactive substance T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}
Given that
Half life of ^{60}_{27} Co, T = 5.3 years.
= 5.3 * 365* 24 * 60 * 60 sec
= 1.67 * 10^8 sec
Strength of the radioactive source R = 8.0 mCi
R = 8.0 * 10^{-3} * 3.7 * 10^{10} decay/sec
R = 29.6 * 10^{7} decay/sec
Step 1: Get an expression for number of radioactive atoms
R = \lambda N
N = \frac{R}{\lambda} ...........................(1)
Half life T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}
\lambda = \frac{\ln 2}{T_{\frac{1}{2}}}
Plug in the \lambda in the equation (1)
N = \frac{R}{\frac{\ln 2}{T_{\frac{1}{2}}}}
N = \frac{R}{\ln 2}*T_{\frac{1}{2}} ..........................(2)
Step 2: Finding the number of radioactive(Co) atoms
Substituting all the known value in the equation (2)
N = \frac{29.6 * 10^{7}}{0.693} * 1.67 * 10^8
N = 7.133 * 10^{16} atoms
Step 3: Obtain the necessary amount(mass) of Co to create a radioactive source
Atomic mass of Cobalt(Co) A = 58.9 \approx 60
Avogadro's number = 6.023 * 10^{23}
Mass of Co atoms = \frac{\text{ Atomic mass * number of radioactive atoms}}{\text{ Avogadro's number }}
=\frac{60 * 7.133 * 10^{16}}{6.023 * 10^{23}}
= 7.12 * 10^{-6} g
Hence, necessary amount(mass) of Co = 7.12 * 10^{-6} g