Question

The half-life period of ${C^{14}}$ is 5670 years. If 56 g of ${C^{14}}$ was initially present, how many atoms of ${C^{14}}$ is left after 22680 years?
A. $1.5 \times {10^{23}}$
B. $2.9 \times {10^{27}}$
C. $1.5 \times {10^{27}}$
D. $2.9 \times {10^{23}}$

Answer 1

We need to use the molar mass and the Avogadro’s constant to find the number of atoms in 56 grams of carbon. Recall that the decay constant is given as the natural log of 2 divided by the half-life.

Formula used: In this solution we will be using the following formulae;
$N = {N_0}{e^{ - \lambda t}}$ where $N$ is the number of atoms left after a time of radioactive decay, ${N_0}$ is the initial number at the start of decay, $\lambda$ is the decay constant of the element, and $t$ is time to decay.
$\lambda = \dfrac{{\ln 2}}{{{T_{1/2}}}}$ where ${T_{1/2}}$ is called the half-life of the element.
$N = \dfrac{m}{M}{N_A}$ where $N$ is the number of atoms in a given mass $m$ of a substance with $M$ molar mass and ${N_A}$ is the Avogadro’s number.

Complete step by step answer:
Generally, the number of atoms left is negatively exponentially dependent on time. This means that the number reduces exponentially with time. The relationship is given as
$N = {N_0}{e^{ - \lambda t}}$ where $N$ is the number of atoms left after a time of radioactive decay, ${N_0}$ is the initial number at the start of decay, $\lambda$ is the decay constant of the element, and $t$ is time to decay.
First. We must know, ${N_0}$.
This can be gotten from the mass as
$N = \dfrac{m}{M}{N_A}$ where $m$ of a substance with $M$ molar mass and ${N_A}$ is the avogadro’s number.
Hence, we have
$N = \dfrac{{56}}{{14}}\left( {6.02 \times {{10}^{23}}} \right) = 2.41 \times {10^{24}}atoms$
Now, we calculate $\lambda$ from
$\lambda = \dfrac{{\ln 2}}{{{T_{1/2}}}}$ where ${T_{1/2}}$ is the half-life
Hence,
$\lambda = \dfrac{{\ln 2}}{{5670}} = 1.2 \times {10^{ - 4}}/s$
Hence, we have
$N = 2.41 \times {10^{24}}{e^{ - 1.2 \times {{10}^{ - 4}} \times 22680}}$
Simply by computation, we have
$N = 1.5 \times {10^{23}}atoms$
Hence, the correct option is A.

Note: Alternatively, without going through lengthy calculations, we could do as follows.
Divide the number of years by the half-life, we have
$\dfrac{{22680}}{{5679}} = 4$
Now, divide the initial value by 2 raised to the power of 4 as in
$N = \dfrac{{{N_0}}}{{{2^4}}} = \dfrac{{2.41 \times {{10}^{24}}}}{{16}} = 1.5 \times {10^{23}}atoms$