A sealed box was found which stated to have contained alloy composed of equal parts by weight of two metals A and B. These metals are radioactive, with half lives of 12 years and 18 years, respectively and when the container was opened it was found to contain 0.53 kg of A and 2.20kg of B. The age of the alloy is M $\times$ 10$^{+n}$ then find M-n.

Hint:Generally decay is nothing but the process by which an unstable atomic nucleus loses energy by radiation. In order to solve this type of question, we need to understand the concept of radioactive decay and the half-life of radioactive substances. By using both the formulas we can solve the age of an alloy.

Complete step by step answer:
Number of radioactive samples of metal A and B.
$
{N_A} = 0.53 \\
\Rightarrow{N_B} = 2.2 \\
$
Given $\dfrac{{{N_B}}}{{{N_A}}} = \dfrac{{2.2}}{{0.53}} = 4.15$
Using the law of radioactivity formula,
$
{N_A} = {N_{0A}}{e^{ - {\lambda _A}t}} \\
\Rightarrow{N_B} = {N_{0B}}{e^{ - {\lambda _B}t}} \\
$
Divide both the above equation
$
\Rightarrow\dfrac{{{N_B}}}{{{N_A}}} = {e^{({\lambda _A} - {\lambda _B})t}}because{N_{0A}} = {N_{0B}} \\
\Rightarrow\dfrac{{{N_B}}}{{{N_A}}} = \dfrac{{{N_{oB}}{e^{ - {\lambda _B}t}}}}{{{N_{0A}}{e^{ - {\lambda _A}t}}}} \\
\Rightarrow t = \dfrac{1}{{({\lambda _A} - {\lambda _B})\ln \left( {\dfrac{{{N_B}}}{{{N_A}}}} \right)}} \\$
From the half-life of radioactive substance
For metal A:
${T_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{{\lambda _A}}}$
$
\Rightarrow{\lambda _A} = \dfrac{{0.693}}{{12}} \\
\Rightarrow{\lambda _A} = 0.05775 year{s^{ - 1}} \\
$
For metal B:
$
{T_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{{\lambda _B}}} \\
\Rightarrow{T_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{18}} \\
\therefore{T_{\dfrac{1}{2}}} = 0.0985 year{s^{ - 1}} \\
$
We find the age of an alloy as 74years. It can be written as 7 $\times 10^{+4}$. Hence M=7, n=4
and M-n = 7 - 4 = 3.

Note:The Law of radioactive decay states that the number of nuclei undergoing decay per unit time is proportional to total number of nuclei in the sample. Half – life of radioactive samples can be expressed as the time in which the number of nuclei samples reduces to one-half of the initial values of the same sample.