Respuesta :
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
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 : 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:
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
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
[tex]k=\frac{2.303}{24\text{ days}}\log\frac{100}{12.5}[/tex]
[tex]k=8.66\times 10^{-2}\text{ days}^{-1}[/tex]
Now we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]8.66\times 10^{-2}=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=8.00\text{ days}[/tex]
Therefore, the half-life of iodine is, 8 days.
