Answer:
The activity is 811.77 Bq
Solution:
As per the question:
Half life of Thorium, [tex]t_{\frac{1}{2}} = 1.405\times 10^{10}\ yrs[/tex]
Mass of Thorium, m = 200 mg = 0.2 g
M = 232 g/mol
Now,
No. of nuclei of Thorium in 200 mg of Thorium:
[tex]N = \frac{N_{o}m}{M}[/tex]
where
[tex]N_{o }[/tex] = Avagadro's number
Thus
[tex]N = \frac{6.02\times 10^{23}\times 0.2}{232} = 5.19\times 10^{20}[/tex]
Also,
Activity is given by:
[tex]\frac{0.693}{t_{\frac{1}{2}}}\times N[/tex]
[tex]A= \frac{0.693}{1.405\times 10^{10}}\times 5.19\times 10^{20} = 2.56\times 10^{10}\ \yr[/tex]
[tex]A = \frac{2.56\times 10^{10}}{365\times 24\times 60\times 60} = 811.77\ Bq[/tex]
