Question

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?

Answer 1

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`

Answer 2

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