Calculate activity of a radioactive substance after 3 hours ?

INPhO 2010
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).

1 Answer
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 #-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)#