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liq [111]
2 years ago
14

The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse

d, what was the initial activity, in microcuries, of the sample?
Chemistry
2 answers:
Umnica [9.8K]2 years ago
5 0
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
padilas [110]2 years ago
5 0

Answer : The initial activity, in microcuries, of the sample was, 120.9 μci

Explanation :

Half-life = 32.5 days

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

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

k=\frac{0.693}{32.5\text{ days}}

k=0.0213\text{ days}^{-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  = 0.0213\text{ days}^{-1}

t = time passed by the sample  = 162.5 days

a = initial amount of the reactant  = ?

a - x = amount left after decay process = 3.8 μci

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

162.5=\frac{2.303}{0.0213}\log\frac{a}{3.8\mu ci}

a=120.9\mu ci

Therefore, the initial activity, in microcuries, of the sample was, 120.9 μci

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