Question

The half-life of mercury-197 is 64.1 hours. If a patient undergoing a kidney scan is given 5.5 ng of mercury-197, how much will remain after 9 days?

Answer 1

t_1/2= Ln2 ---> =64.1/Ln2 --->= 92.5

= 1/   ---->= 0.0108

N(t)=N⁰e^-t

N(t)=number of atoms on the sample at the time "t"

N0=initial number of atoms on the sample

=desintegration constant

atomic mass= 197g/mol -----> 1mol Hg-197 -------- 197g

                                            x=2.8 x 10^-11mol -------5.5x 10^-9g

1 mol Hg ------ 6.02 x 10²³ atoms

2.8 x 10^-11mol -----x= 1.7 x 10¹³ atoms Hg

N(t)=N⁰e^-t ---> N(t)=1.7 x 10¹³ e^{-0.0108 x 216hours} ---> N(t)= 1.65 x 10¹² atoms Hg ----> 2.74 x 10^-12 mol this is the answer. After 9 days you will have 2.74 x 10^-12 mol of Hg, or 2.74 x 10^-12 mol x 197g/mol= 5.4 x 10^-10g