Question

The half-life of thallium-201 is 73 hours. How many hours will it take for an amount of thallium-201 to decay so that only 5% of the original amount remains?

Answer 1

Answer:

132 days

Step-by-step explanation:

Half life of Thallium-201 = 73 hours

i.e. in 73 hours it decomposes to 1/2 of its initial quantity

In next 73 hours it will be 1/4

If initial quantity is $N_0$

Then 5% of initial amount

$=0.05N_0$

Now,

$0.05N_0=N_0\times (\frac{1}{2})^n$

or, $2^n=\frac{1}{0.05}$

or, $2^n=20$

or, $n\log 2=\log 20$

$\implies n=\frac{\log 20}{\log 2}$

$\implies n=\frac{\log (10\times 2)}{\log 2}$

$\implies n=\frac{(1+\log 2)}{\log 2}$

$\implies n=\frac{1.3010}{0.3010}$

$\implies n=4.32$

Therefore, the time taken

$=t_{1/2}\times 4.32$ hours

$=73\times 4.32$ hours

$=3166.56$ hours

$=\frac{3166.56}{24}$ days

$=131.94$ days

≈ 132 days

Hope this helps.