Question

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).

Answer 1

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 `-dN` expected to occur in an infinitesimal interval of time `dt` is proportional to the number of atoms `N` present in the sample. Mathematically,

`- (d N) /(d t)prop N`
`(d N) /(d t)-=A=lambda N` ........(1)
where `A` is activity and `lambda` is proportionality constant also called decay constant.

Alternatively the probability of decay `-(dN)/N` is proportional to incremental of time `dt`.

`=>- (d N) /Nprop 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_(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(t)=N_0(1/2)^(t/t_(1/2))` .......(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`
`=>t_(1/2)=ln2/lambda` .........(5)

It is given that sample decays via two exponential decay processes simultaneously. Therefore, the actual half-life `T_(1/2)` can expressed in terms of two half-lives `t_1 and t_2` as

`1/ T_(1 / 2) = 1/ t_1 + 1/ t_2` ..........(6)

Two half lives are given as `t_1=6\ hrandt_2=2.12xx10^5\ y`. Inserting these values in (6) we see that second term on the RHS is `"<<"` compared to the first term. Therefore, effective half-life will be `=6\ hr`, as already given in the question.

(a) Inserting given value in (5)

`6=ln2/lambda_k`
`=>lambda_k=ln2/6`
`=>lambda_k=0.1155\ hr^-1`

(b) Initial activity is given `A_0=1.0\ muCi`, where `A_0=lambdaN_0` [from (1).]

From (3) number of nuclei remaining after `3\ hr`

`N(3\ hr)=N_0\ e ^ (-0.1155xx3)`
`N(3\ hr)=0.707N_0`

From (1)

`A(3\ hr)=0.707A_0`
`A(3\ hr)=0.707\ muCi`

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

Inserting various values in (1) we get `N_0` as

`1.0xx10^-6xx(3.70xx10^10)=0.1155 N_0`
`=> N_0=(1.0xx10^-6)/0.1155xx(3.70xx10^10)`