Question

5/3if 87.5 percent of a sample of pure 131 i decays in 24 days, what is the half-life of 131 i?

Answer 1

1st time of decay: 50%
2nd time of decay: 50%+25%=75%
3rd time of decay: 75+12.5%=87.5%
Therefore, 87.5% of the samples decays in 3 half-lives
Therefore, 24 days is 3 half-lives
Therefore:
1 half-life is 24/3 = 8 days

Answer 2

Answer : The half-life of iodine is, 8 days.

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

`k=(2.303)/(t)\log(a)/(a-x)`

where,

k = rate constant = ?

t = time passed by the sample = 24 days

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = 100 - 87.5 = 12.5 g

Now put all the given values in above equation, we get

`k=\frac{2.303}{24\text{ days}}\log(100)/(12.5)`

`k=8.66* 10^(-2)\text{ days}^(-1)`

Now we have to calculate the rate constant, we use the formula :

`k=(0.693)/(t_(1/2))`

`8.66* 10^(-2)=(0.693)/(t_(1/2))`

`t_(1/2)=8.00\text{ days}`

Therefore, the half-life of iodine is, 8 days.