Question

The half-life of a radioactive isotope is ${\text{3 hours}}$. If the initial mass of the isotope were ${\text{300g}}$, the mass of it remaining undecayed after ${\text{18 hours}}$ would be:
A.${\text{2}}{\text{.34gm}}$
B.${\text{1}}{\text{.17gm}}$
C.${\text{9}}{\text{.36gm}}$
D.${\text{4}}{\text{.68gm}}$

Answer 1

To answer this question, you should recall the concept of integrated rate law. It is used to provide a relationship between the rate of the reaction and the concentrations of the reactants participating in it. It can be only determined experimentally and not from the balanced chemical equation. It is related to the half life of a reaction by the formula given below.
The formula used: ${\text{N}} = {{\text{N}}_0}{\left( {\dfrac{1}{2}} \right)^n}\;$where ${\text{N}}$=Amount of substance after n half lives and ${{\text{N}}_{\text{0}}}$=Amount of substance initially.

Complete step by step answer:
$n$= number of half lives can be calculated by dividing the given time by the half-life$= \dfrac{{18}}{3} = 6$ .
Now substituting this in the formula for remaining radioactive substance: ${\text{N}} = 300 \times {\left( {\dfrac{1}{2}} \right)^6} = 4.68{\text{g}}\;\;\;$
Hence the mass remaining undecayed is ${\text{4}}{\text{.68gm}}$.

Hence option D is correct.

Note:
The overall order of any reaction is the sum of the partial components of a chemical reaction. The rate$= {\text{ }}k{\left[ A \right]^x}{\left[ B \right]^y}\;$then the overall order can be written as $x + y$. As we said that the order provides a relationship between rate and concentration of components:
If the reaction is a zero-order reaction, change in the reactant concentration will not affect the reaction rate.
In the case of the first order, doubling the reactant concentration will double the reaction rate.
In the case of second-order, doubling the concentration of the reactants will quadruple the overall reaction rate.
In the case of third-order, the overall rate increases by eight times when the reactant concentration is doubled. The important formula for the half-life of a reaction which varies with the order of the reaction.
In the case of a zero-order reaction, the expression for the half-life is: ${t_{1/2}}\; = {\text{ }}\dfrac{{{{\left[ R \right]}_0}}}{{2k}}$
In the case of the first-order reaction, the expression for the half-life is given by: ${t_{1/2}}\; = \dfrac{{{\text{ }}0.693}}{k}$
In case of a second-order reaction, the expression for the half-life of the reaction is${t_{1/2}} = \;\dfrac{1}{{k{{[R]}_0}}}$