Question

If the half-life of technitium-104 is 18.0 min, how much of a 14.2-g sample will remain after 72 minutes?

Answer 1

Answer : The correct option is, 0.888 g

Solution : Given,

As we know that the radioactive decays follow first order kinetics.

First we have to calculate the rate constant of a radioisotope.

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

$18min=\frac{0.693}{k}$

$k=0.0385min^{-1}$

Now we have to calculate the amount remains after 72 minutes.

The expression for rate law for first order kinetics is given by :

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

where,

k = rate constant  = $0.0385min^{-1}$

t = time taken for decay process  = 72 min

a = initial amount of the reactant  = 14.2 g

a - x = amount left after decay process  = ?

Putting values in above equation, we get the value of amount left.

$0.0385min^{-1}=\frac{2.303}{72min}\log\frac{14.2g}{a-x}$

$a-x=0.888g$

Therefore, the amount remain after 72 min will be, 0.888 g

Answer 2

The decay of a radioactive isotope can be predicted using the formula: A = Ao[2^(-t/T_0.5)] where A is the amount after time t, Ao is the original amount and T_0.5 is the half-life. Using the equation and the given values, 0.888 g of the sample will remain after 72 minutes.