# Calculate activity of a radioactive substance after 3 hours ?

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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).

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

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

#- (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

#=>- (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

#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

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

Two half lives are given as

(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

From (3) number of nuclei remaining after

#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

#1.0xx10^-6xx(3.70xx10^10)=0.1155 N_0#

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