# Calculate activity of a radioactive substance after 3 hours ?

## Please help me solve this , The 1ml Blood part of the question is for some further parts and not related to part (a) and (b).

Mar 5, 2018

This is what I get

#### Explanation:

For decay process of an unstable nucleus is entirely random. It is impossible to predict when a specific atom will decay. However, we talk about probability of decay of a particular nucleus at a given instant in time.
Therefore, in a given sample of radioactive material, the number of decay events $- \mathrm{dN}$ expected to occur in an infinitesimal interval of time $\mathrm{dt}$ is proportional to the number of atoms $N$ present in the sample. Mathematically,

$- \frac{d N}{d t} \propto N$
$\frac{d N}{d t} \equiv A = \lambda N$ ........(1)
where $A$ is activity and $\lambda$ is proportionality constant also called decay constant.

Alternatively the probability of decay $- \frac{\mathrm{dN}}{N}$ is proportional to incremental of time $\mathrm{dt}$.

$\implies - \frac{d N}{N} \propto \mathrm{dt}$
=>- (d N)/ N = λ d t  .....(2)
The negative sign indicates that $N$ decreases with increase of time.

This first-order differential equation has a solution of the form

N ( t ) = N_0\ e ^ (-λt) .......(3)
where ${N}_{0}$ is the value of $N$ at time $t = 0$.

Here we introduce term half life ${t}_{\frac{1}{2}}$which is time in which on an average exactly half of the nuclei would have decayed. Mathematically it can be written as

$N \left(t\right) = {N}_{0} {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\frac{1}{2}\right)}$ .......(4)

Comparing (3) and (4) we get

(1/2)^(t/t_(1/2))=e ^ (-λt)

Tanking natural logarithm of both sides

ln(1/2)^(t/t_(1/2))=lne ^ (-λt)
=>(t/t_(1/2))ln(1/2)= (-λt)lne
$\implies {t}_{\frac{1}{2}} = \ln \frac{2}{\lambda}$ .........(5)

It is given that sample decays via two exponential decay processes simultaneously. Therefore, the actual half-life ${T}_{\frac{1}{2}}$ can expressed in terms of two half-lives ${t}_{1} \mathmr{and} {t}_{2}$ as

$\frac{1}{T} _ \left(\frac{1}{2}\right) = \frac{1}{t} _ 1 + \frac{1}{t} _ 2$ ..........(6)

Two half lives are given as ${t}_{1} = 6 \setminus h r \mathmr{and} {t}_{2} = 2.12 \times {10}^{5} \setminus y$. Inserting these values in (6) we see that second term on the RHS is $\text{<<}$ compared to the first term. Therefore, effective half-life will be $= 6 \setminus h r$, as already given in the question.

(a) Inserting given value in (5)

$6 = \ln \frac{2}{\lambda} _ k$
$\implies {\lambda}_{k} = \ln \frac{2}{6}$
$\implies {\lambda}_{k} = 0.1155 \setminus h {r}^{-} 1$

(b) Initial activity is given ${A}_{0} = 1.0 \setminus \mu C i$, where ${A}_{0} = \lambda {N}_{0}$ [from (1).]

From (3) number of nuclei remaining after $3 \setminus h r$

$N \left(3 \setminus h r\right) = {N}_{0} \setminus {e}^{- 0.1155 \times 3}$
$N \left(3 \setminus h r\right) = 0.707 {N}_{0}$

From (1)

$A \left(3 \setminus h r\right) = 0.707 {A}_{0}$
$A \left(3 \setminus h r\right) = 0.707 \setminus \mu C i$

.-.-.-.-.-.-.-.-.-.-.-
Specifically

Inserting various values in (1) we get ${N}_{0}$ as

$1.0 \times {10}^{-} 6 \times \left(3.70 \times {10}^{10}\right) = 0.1155 {N}_{0}$
$\implies {N}_{0} = \frac{1.0 \times {10}^{-} 6}{0.1155} \times \left(3.70 \times {10}^{10}\right)$