Answer:
Effective half-time of the tracer is 3.6 days
Explanation:
The formula for calculating the decay due to excretion for the first process is ;
[tex]\frac{dN_1}{dt } = - \lambda _e N_o[/tex]
here ;
[tex]N_o[/tex] = initial number of tracers
Then to the second process ; we have :
[tex]\frac{dN_2}{dt } = - \lambda _e N_o[/tex]
The total decay is as a result of the overall process occurring ; we have :
[tex]\frac{dN_{total}}{dt } = \frac{dN_1}{dt} + \frac{dN_2}{dt}[/tex] ------ (1)
here ;
[tex]\frac{dN_{total}}{dt } = \lambda _{total} N_o[/tex]
Putting the values in (1);we have :
[tex]- \lambda _{Total} N_o = - \lambda_e N_o + ( -\lambda r N_o})[/tex]
[tex]\lambda _{Total} = \lambda_e + \lambda r[/tex]
As we also know that:
[tex]\frac{1}{t_{1/2}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}[/tex]
[tex]\frac{1}{t_{1/2}}_{effective}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}[/tex]
[tex]\frac{1}{t_{1/2}}_{effective}} = \frac{9+6}{9*6}[/tex]
[tex]\frac{1}{t_{1/2}_{effective}}}=\frac{15}{54}[/tex]
[tex]t_{1/2}_{effective}} = \frac{54}{15}[/tex]
= 3.6 days
Answer:
The effective half-life is [tex](t_{\frac{1}{2} })_T = 8.6 \ days[/tex]
Explanation:
From the question we are told that
The excretion half-life is [tex]t__{\frac{1}{2} } excretion} = 6 \ days[/tex]
The radiation half-life is [tex]t__{\frac{1}{2} } radiation} = 9 \ days[/tex]
The decay due to excretion is mathematically represented as
[tex]\frac{dN_a}{dt} = \lambda_e N_i[/tex]
Where [tex]Ni[/tex] is the original number of tracers
The decay due to excretion is mathematically represented as
[tex]\frac{dN_b}{dt} = \lambda_r N_i[/tex]
Now from the question we are total decay is as result of the combined decay of both processes
We have that
[tex]\frac{dN_T}{dt} =\lambda_T N_i= \frac{dN_a}{dt} + \frac{dN_b}{dt}[/tex]
Substituting for the formula above
[tex]\lambda_T = \lambda _e + \lambda_r[/tex]
Generally the formula for half-life is
[tex]t__{\frac{1}{2} }}= \frac{0.693}{\lambda }[/tex]
So [tex]\lambda = \frac{0.693}{t_{\frac{1}{2} }}[/tex]
Substituting this into the above equation
[tex][\frac{0.693}{t_\frac{1}{2} } ]_T =[\frac{0.693}{t_\frac{1}{2} } ]_e + [\frac{0.693}{t_\frac{1}{2} } ]_r[/tex]
[tex][\frac{1}{t_\frac{1}{2} } ]_T =\frac{[\frac{1}{t_\frac{1}{2} } ]_e + [\frac{1}{t_\frac{1}{2} } ]_r}{[\frac{1}{t_\frac{1}{2} } ]_e * [\frac{1}{t_\frac{1}{2} } ]_r}[/tex]
Substituting values
[tex][\frac{1}{t_\frac{1}{2} } ]_T =\frac{ 9+6}{9 * 6}[/tex]
[tex][\frac{1}{t_\frac{1}{2} } ]_T =\frac{ 15}{54}[/tex]
[tex](t_{\frac{1}{2} })_T =\frac{ 54 }{15}[/tex]
[tex](t_{\frac{1}{2} })_T = 8.6 \ days[/tex]
