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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

I found an answer from content.wolfram.com

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.

For more information, see Weak Interaction

I found an answer from ehs.stanford.edu

**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