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When radioactive tracers are used for medical scans, the tracers both decay in the body via radiation, as well as being excreted from the body. The effective half-life of the radioactive tracer is due to the combination of both effects. Medical experiments show that a stable (nonradioactive) isotope of a particular element have an excretion half-life of 6 days. A radioactive isotope of the same element has a half-life of 9 days. What is the effective half-life of the radioactive tracer due to both effects?

User Fmstrat
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Answer:

The effective half-life is
(t_{(1)/(2) })_T = 8.6 \ days

Step-by-step explanation:

From the question we are told that

The excretion half-life is
t__{(1)/(2) } excretion} = 6 \ days

The radiation half-life is
t__{(1)/(2) } radiation} = 9 \ days

The decay due to excretion is mathematically represented as


(dN_a)/(dt) = \lambda_e N_i

Where
Ni is the original number of tracers

The decay due to excretion is mathematically represented as


(dN_b)/(dt) = \lambda_r N_i

Now from the question we are total decay is as result of the combined decay of both processes

We have that


(dN_T)/(dt) =\lambda_T N_i= (dN_a)/(dt) + (dN_b)/(dt)

Substituting for the formula above


\lambda_T = \lambda _e + \lambda_r

Generally the formula for half-life is


t__{(1)/(2) }}= (0.693)/(\lambda )

So
\lambda = \frac{0.693}{t_{(1)/(2) }}

Substituting this into the above equation


[(0.693)/(t_(1)/(2) ) ]_T =[(0.693)/(t_(1)/(2) ) ]_e + [(0.693)/(t_(1)/(2) ) ]_r


[(1)/(t_(1)/(2) ) ]_T =([(1)/(t_(1)/(2) ) ]_e + [(1)/(t_(1)/(2) ) ]_r)/([(1)/(t_(1)/(2) ) ]_e * [(1)/(t_(1)/(2) ) ]_r)

Substituting values


[(1)/(t_(1)/(2) ) ]_T =( 9+6)/(9 * 6)


[(1)/(t_(1)/(2) ) ]_T =( 15)/(54)


(t_{(1)/(2) })_T =( 54 )/(15)


(t_{(1)/(2) })_T = 8.6 \ days

User HammerFet
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Answer:

Effective half-time of the tracer is 3.6 days

Step-by-step explanation:

The formula for calculating the decay due to excretion for the first process is ;


(dN_1)/(dt ) = - \lambda _e N_o

here ;


N_o = initial number of tracers

Then to the second process ; we have :


(dN_2)/(dt ) = - \lambda _e N_o

The total decay is as a result of the overall process occurring ; we have :


(dN_(total))/(dt ) = (dN_1)/(dt) + (dN_2)/(dt) ------ (1)

here ;


(dN_(total))/(dt ) = \lambda _(total) N_o

Putting the values in (1);we have :


- \lambda _(Total) N_o = - \lambda_e N_o + ( -\lambda r N_o})


\lambda _(Total) = \lambda_e + \lambda r

As we also know that:


(1)/(t_(1/2)) = ([t_(1/2)]_(radiation)+[t_(1/2)]_(excretion))/([t_(1/2)]_(radiation)*[t_(1/2)]_(excretion))


(1)/(t_(1/2))_(effective)} = ([t_(1/2)]_(radiation)+[t_(1/2)]_(excretion))/([t_(1/2)]_(radiation)*[t_(1/2)]_(excretion))


(1)/(t_(1/2))_(effective)} = (9+6)/(9*6)


(1)/(t_(1/2)_(effective))}=(15)/(54)


t_(1/2)_(effective)} = (54)/(15)

= 3.6 days

User Mahdi
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