Respuesta :
Answer : The original activity will be, 303 millicuries.
Explanation :
Half-life = 15 hr
First we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{15hr}[/tex]
[tex]k=4.62\times 10^{-2}\text{ hr}^{-1}[/tex]
Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]4.62\times 10^{-2}\text{ hr}^{-1}[/tex]
t = time passed by the sample = 24 hr
a = initial amount of the reactant = ?
a - x = amount left after decay process = 100 millicuries
Now put all the given values in above equation, we get
[tex]24=\frac{2.303}{4.62\times 10^{-2}}\log\frac{a}{100}[/tex]
[tex]a=302.97\text{ millicuries}\approx 303\text{ millicuries}[/tex]
Therefore, the original activity will be, 303 millicuries.
