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The half –life for the radioactive decay of 32P is 14.3 days. How many days would be required for a couple of a radiopharmaceutical containing 32P to decrease to 20% of its initial activity?

1 Answer

5 votes

m=0,2m*((1)/(2))^{(t)/(14,3)} \ \ | \:0,2m\\\\ (m)/(0,2m)=((1)/(2))^{(t)/(14,3)}\\\\ 5=((1)/(2))^{(t)/(14,3)} \\\\ log_(a)b=c \ \ \Leftrightarrow \ \ a^(c)=b\\\\ (t)/(14,3)=log_(1)/(2)5\\\\ (t)/(14,3)=2,31 \ \ |*14,3\\\\ t = 33,033\approx 33days
answered
User Luuk Krijnen
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