Question

The piece of wood from a recently cut tree shows $20$ decays per minute. A wooden piece of the same size placed in a museum (obtained from a tree cut many years back) shows $2$ decays per minute. If half life of ${C^{14}}$ is $5730$ years, then age of the wooden piece placed in the museum
is approximately
A. $10439\,years$
B. $13094\,years$
C. $19039\,years$
D. $39040\,years$

Answer 1

Hint: Half-life related to a radioactive material is the time period needed for the decay of one-half of the atomic nuclei in a radioactive sample.
We can use the formula to calculate the time period,
 $\dfrac{A}{{{A_{\text{o}}}}}{\text{ = }}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{\dfrac{t}{{{T_{\dfrac{1}{2}}}}}}}$
where ${A_{\text{o}}}$ refers to the starting quantity of the isotope that will decay, $A$ is the remaining quantity, ${{\text{T}}_{{\text{1/2}}}}$represents half-life of the decaying quantity and ${\text{t}}$ is the time taken.

Complete step by step answer:
Half-life as the name denotes means half of a particular substance to react chemically.
It is a common term in nuclear physics and the survival of atoms in radioactive decay depends on the half-life. Half-life can encourage the classification of any form of decay whether rapidly changing or not.
Radioactive elements emit energy all the time, and their lifespan is determined by half-life — how long it takes for half of a specified quantity to emit its radiation and degrade into another substance.
Half-life is the time required for a quantity to be reduced to half of its original value. The concept is widely used in nuclear physics to explain how easily unstable atoms are undergoing or how long stable atoms have survived radioactive decay.
Given,
Half-life of the radioactive substance, $ = 5730\,years$
Initial quantity, ${A_ \circ } = 20$ decays per minute
Remaining quantity, $A = 2$ decays per minute
We know that,
$
  A = {A_ \circ }{e^{ - \lambda t}} \\
  \Rightarrow 2 = 20{e^{ - \lambda t}} \\
 $

$ \Rightarrow t = \dfrac{{\log 10}}{\lambda }$ ...... (i)
Also,
$\lambda = \dfrac{{\log 2}}
{{{t_{\left( {\dfrac{1}
{2}} \right)}}}}$

Putting the above value in equation (i), we get:
$ t = \dfrac{{\log 10}}
{{\dfrac{{\log 2}}
{{{t_{\dfrac{1}
{2}}}}}}} \\
\Rightarrow t = \dfrac{{\log 10}}
{{\dfrac{{\log 2}}
{{5730}}}} \\
\Rightarrow t = 19039\,years \\
$
Hence, option C is correct.

Thus, the age of the wooden piece placed in the museum is approximately $19039\,years$ if half life of ${C^{14}}$ is $5730$ years.

Note: There may be many half-lives depending on the reaction. A radio-active substance may have first half-life, second half-life etc and we have to change the calculations based on that. A specific radioactive material has a steady half-life.