Question

The isotope 208TI undergoes B decay with a half-life of 3.1 min. How long will it take for 99.0% of a sample of pure 208 TI to decay?

Answer 1

Answer:

20.9 min time taken forn 99.0% of a sample of pure $^{208}Tl$.

Explanation:

From the given,

half line of $^{208}Tl$ = 3.1 min

$t_{1/2}= 3.1min$

$\lambda =\frac{ln(2)}{t_{1/2}}$

$\lambda =\frac{ln(2)}{3.1} =0.22min^{-1}$

The ratio of the initial amount $N_{o}$ and amount in time is calculated as follows.

$\frac{N}{N_{o}}=e^{-\lambda t}$

Now 99% got delayed, the left amount will be 1%.

$\frac{N}{N_{o}}=\frac{1}{100}$

$0.01=e^{-0.22min^{-1}.t}$

$ln(0.01)=(-0.22min^{-1}.t)$

$t=\frac{-ln(0.01)}{-0.22min^{-1}}=20.9min$

Therefore, 20.9 min time taken forn 99.0% of a sample of pure $^{208}Tl$.