Question

A piece of wood from a recently cut tree shows 20 decay per minute. A wooden piece of the same size placed in a museum (obtained from a tree cut years back) shows 2 decay per minute. If half-life of ${C^{14}}$ is 5730 years, then the age of the wooden piece in the museum is approximately:
(A) 10439 years
(B) 13094 years
(C) 19039 years
(D) 39049 years

Answer 1

Hint : The time it takes to start decaying at a particular rate after first decaying at one rate would be the age of the substance. Half-life can be defined as the time it will take half of the atom in a particular substance to decay. The ratio of the initial rate of decay to the final rate of decay is equal to the ratio of the initial number of constant to the final number of constant.

Formula used: In this solution we will be using the following formula;
$N = {N_0}{e^{ - \dfrac{{In2}}{{{T_{\dfrac{1}{2}}}}}t}}$ where $N$ is the final number of atoms at a particular time $t$ , ${N_0}$ is the initial number of atoms at time $t = 0$ , and ${T_{\dfrac{1}{2}}}$ is half-life of the substance.

Complete step by step answer:
To solve for the age, we must recall the formula
$N = {N_0}{e^{ - \dfrac{{\ln 2}}{{{T_{\dfrac{1}{2}}}}}t}}$ where $N$ is the final number of atoms at a particular time $t$ , ${N_0}$ is the initial number of atoms at time $t = 0$ , and ${T_{\dfrac{1}{2}}}$ is half-life of the substance.
Hence, $\dfrac{N}{{{N_0}}}{ = _0}{e^{ - \dfrac{{\ln 2}}{{{T_{\dfrac{1}{2}}}}}t}}$
However, the rate of decay of atoms was given, not the number of atoms. But
$\dfrac{N}{{{N_0}}} = \dfrac{{\dfrac{{dN}}{{dt}}}}{{\dfrac{{{N_0}}}{{dt}}}}$
Then,
$\dfrac{N}{{{N_0}}} = \dfrac{{\dfrac{{dN}}{{dt}}}}{{\dfrac{{{N_0}}}{{dt}}}} = \dfrac{2}{{20}} = \dfrac{1}{{10}}$
By inserting all known values
$\dfrac{1}{{10}} = {e^{ - \dfrac{{\ln 2}}{{5730}}t}}$
Taking the natural log of both sides
$\ln \dfrac{1}{{10}} = \ln {e^{ - \dfrac{{\ln 2}}{{5730}}t}}$
$\Rightarrow - \ln 10 = - \dfrac{{\ln 2}}{{5730}}t$
By making time subject of formula, we have
$t = \dfrac{{\ln 10 \times 5730}}{{\ln 2}} = \dfrac{{2.303 \times 5730}}{{0.693}}$
$\Rightarrow t = 19039{\text{ years}}$
Hence, the correct option is C.

Note:
Alternatively, without the use of formula, we can use this reasoning. Half-life is the time it takes half of the atom of a radioactive substance to decay, i.e. the number of atoms left will be the number of atoms it began with. Now since the number of atoms is also proportional to the rate, then the wood after 5730 years should have a rate of 10 decay per minute, and after another 5730 years, it should decay to half of that which is 5 decay per minute. After another 5730 years, it decays to 2.5 decay per minute. Adding the years together we have, 17190 years. Hence, observing the option by 2 decay per minute, the only reasonable answer should be 19039 years.