Question

0% of its original value? View Available Hint(s) If a radioactive isotope has a half-life of 26.5 days, how many days does it take for a sample of the isotope to decrease by 35.0% of its original value? 40.1 days 16.5 days 7.15 days 0.0163 days

Answer 1

Answer:

16.5 days

Explanation:

Given that:

Half life = 26.5 days

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,  

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{26.5}\ days^{-1}$

The rate constant, k = 0.02616 days⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

35.0 % is decomposed which means that 0.35 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 1 - 0.35 = 0.65

t = 7.8 min

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.65=e^{-0.02616\times t}$

t = 16.5 days.