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?

3.55g

1.78g

0.888g

0.444g

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)=(0.693)/(k)`

`18min=(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=(2.303)/(t)\log(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)=(2.303)/(72min)\log(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.