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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