Question

A sample of radiosodium () has a half-life of 15 hr. If the sampleâs activity is 100 millicuries after 24 hr, approximately what must its original activity have been?

Answer 1

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 :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{15hr}$

$k=4.62\times 10^{-2}\text{ hr}^{-1}$

Now we have to calculate the time passed.

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

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $4.62\times 10^{-2}\text{ hr}^{-1}$

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

$24=\frac{2.303}{4.62\times 10^{-2}}\log\frac{a}{100}$

$a=302.97\text{ millicuries}\approx 303\text{ millicuries}$

Therefore, the original activity will be, 303 millicuries.