Question

A radioactive isotope has a half – life of 10 days. If today 125 mg is left over, what was its original weight 40 days earlier?
A) 2g
B) 600mg
C) 1g
D) 1.5g

Answer 1

Apply the meaning of half-life and the number of half-lives that will happen in 40 days to equate the amount 40 days earlier with the amount 40 days after.

Complete answer:
In radioactive processes, particles or electromagnetic radiation are emitted from the nucleus. The most common forms of radiation emitted have been traditionally classified as $alpha(\alpha ),beta\left( \beta \right),$ and $gamma\left( \gamma \right)$radiation. Nuclear radiation occurs in other forms, including the emission of protons or neutrons or spontaneous fission of a massive nucleus.
Radioactivity is a first-order process in which half-life is independent of the initial amount of substance, that is, whichever may be the initial amount it will reduce to half its original value in the same time. This time is symbolically referred to as ${t_{1/2}}$ .
In the present case, half-life is 10 days, i.e., the amount reduces to half its original value in 10 days. In 40 days, $\dfrac{{40}}{{10}} = 4$ such reductions to half will happen successively. That is, if the original amount is 100, it will reduce to $\dfrac{1}{2} \times 100 = 50$ in 10 days, and to ${\left( {\dfrac{1}{2}} \right)^2} \times 100 = 25$ in 20 days from the start or 2 half-lives, and so on.
That is, if the original amount was ${N_0}$, the present amount of N can be deduced by the following equation:
$N = {N_0} \times {\left( {\dfrac{1}{2}} \right)^n}$; where $n$ is the number of half-lives.
Here, $N = 125mg$ and n = 4
Therefore, the equation becomes:
$125mg = {N_0} \times {\left( {\dfrac{1}{2}} \right)^4}$
$\Rightarrow {N_0} = \left( {125 \times 16} \right) = 2000mg = 2g$

Hence, option A 2g is the correct answer.

Note: Do not apply the first-order integrated rate-law as it will involve finding decay-constant, $k$, from the half-life and then filling it in a logarithmic equation making the process error-prone. Instead, the application of simple logic as shown above solves the problem in no time.