Question

If the half-lives of a radioactive element for \[\alpha \]- decay and $\beta $- decay are 4 years and 12 years respectively. What percent would its total activity be of its initial activity after 12 years?
${\text{A}}{\text{.}}$ 50
${\text{B}}{\text{.}}$ 25
${\text{C}}{\text{.}}$ 12.5
${\text{D}}{\text{.}}$ 6.25

Answer 1

Hint: Here, we will proceed by finding out the effective time corresponding to the half live remaining of the radioactive element. Half life is the time required to decay one half of the atomic nuclei. Then, finally we will find out the activity of the radioactive element after 12 years.

Complete answer:
Given, Half-live of a radioactive element for \[\alpha \]- decay ${{\text{T}}_\alpha }$ = 4 years
Half-live of a radioactive element for $\beta $- decay ${{\text{T}}_\beta }$ = 12 years
As we know that the effective time is given by
$
  \dfrac{1}{{{{\text{T}}_{{\text{eff}}}}}} = \dfrac{1}{{{{\text{T}}_\alpha }}} + \dfrac{1}{{{{\text{T}}_\beta }}} \\
   \Rightarrow \dfrac{1}{{{{\text{T}}_{{\text{eff}}}}}} = \dfrac{1}{4} + \dfrac{1}{{12}} \\
 $
Taking 12 as the LCM on the RHS of the above equation, we get
$
   \Rightarrow \dfrac{1}{{{{\text{T}}_{{\text{eff}}}}}} = \dfrac{{3 + 1}}{{12}} \\
   \Rightarrow \dfrac{1}{{{{\text{T}}_{{\text{eff}}}}}} = \dfrac{4}{{12}} \\
   \Rightarrow \dfrac{1}{{{{\text{T}}_{{\text{eff}}}}}} = \dfrac{1}{3} \\
 $
By cross multiplying the above equation, we get
$ \Rightarrow {{\text{T}}_{{\text{eff}}}} = 3$ years
So, the effective half-live of the radioactive element is 3 years. This means that the radioactive element becomes half of the present quantity after every 3 years.
Let N be the initial quantity of the radioactive element. After 3 years, half of the initial quantity of the radioactive element i.e., \[\dfrac{{\text{N}}}{2}\] will be left. After another 3 years (i.e., after 6 years), half of the left quantity of the radioactive element i.e., half of \[\dfrac{{\text{N}}}{2}\] will be left which is equal to \[\dfrac{{\text{N}}}{4}\]. After another 3 years (i.e., 9 years), half of \[\dfrac{{\text{N}}}{4}\] will be left which is \[\dfrac{{\text{N}}}{8}\]. After another 3 years (i.e., 12 years), half of \[\dfrac{{\text{N}}}{8}\] will be left which is \[\dfrac{{\text{N}}}{{16}}\]

Initial quantity of the radioactive element = N
Final quantity of the radioactive element left after 12 years = \[\dfrac{{\text{N}}}{{16}}\]
Percentage of the total activity of its initial activity after 12 years = $\dfrac{{{\text{Final quantity of the radioactive element left after 12 years}}}}{{{\text{Initial quantity of the radioactive element}}}} \times 100$
$ \Rightarrow $ Percentage of the total activity of its initial activity after 12 years = $\dfrac{{\dfrac{{\text{N}}}{{16}}}}{{\text{N}}} \times 100 = \dfrac{{100}}{{16}} = 6.25$ percent
Therefore, 6.25 percent would be its initial activity after 12 years.

So, the correct answer is “Option D”.

Note:
In this particular problem, it is important to know the basic difference between the \[\alpha \]- decay and $\beta $- decay of any radioactive element. In \[\alpha \]- decay, an alpha particle is emitted by the nucleus of the radioactive element whereas in $\beta $- decay, a beta particle is emitted by the nucleus of the radioactive element.