Question

If a radioactive substance reduces to \[\dfrac{1}{{16}}\] of its original mass in \[40\] days, what is its half-life?
A. \[10\] days
B. \[20\] days
C. \[40\] days
D. None of these

Answer 1

Hint: In the given question, we have to determine the half-life of a radioactive substance. For this, we will use the formula of activity of a radioactive substance at any time \[t\], that is the amount of substance that will decay.

Formula used:
The amount of substance that will decay is given by
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^{t/T_{1/2}}}\]
Here, \[{N_0}\] is the amount of substance that will initially decay and \[N\] is the quantity that still remains and its decay has not taken place after a time \[t\].

Complete step by step solution:
We know that the amount of substance that will decay is,
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^{t/T_{1/2}}} \\ \]
Here, \[t = 40\] days
This gives, \[\left( {\dfrac{1}{{16}}} \right) = {\left( {\dfrac{1}{2}} \right)^{40/T_{1/2}}} \\ \]
By simplifying, we get
\[{\left( {\dfrac{1}{2}} \right)^4} = {\left( {\dfrac{1}{2}} \right)^{40/T_{1/2}}} \\ \]
Hence, by comparing the power, we get
\[4 = 40/T_{1/2}\]
By simplifying, we get
\[T_{1/2} = 10\]
Thus, if a radioactive substance reduces to \[\dfrac{1}{{16}}\] of its original mass in \[40\] days, then its half-life is \[10\] days.

Therefore, the correct option is A.

Additional Information: An atom's nucleus shows radioactivity as a result of nuclear instability. Henry Becquerel made this discovery. The phenomenon of radioactivity takes place when an unstable atom's nucleus releases radiation to lose energy. The term half-life is defined as the time taken for half of the atoms in a radioactive material to undergo decay.

Note: Many students make mistakes in writing the formula of activity of a radioactive substance at any time \[t\]. Also, it is necessary to simplify it correctly to get the correct half-life.