Question

Radioactive carbon-14, in a wood sample, decays with a half-life of 5700 years. The fraction of the radioactive carbon-14, that remains after a decay period of $17,100$ year is?

Answer 1

Hint: Recall the relationship between the initial amount of radioactive substance and the final amount after a certain time t. In other words, we know that this is an exponential relation. Use this relation to obtain the final amount in terms of the half-life period by using the fact that the quantity remaining at the half-life is half the initial quantity of the radioactive substance. Substitute the given values and solve this arithmetically to arrive at the appropriate result.

Formula Used: Exponential radioactive decay: $N = N_0\left(\dfrac{1}{2}\right)^{\dfrac{t}{t_{1/2}}}$

Complete step-by-step solution:
We know that radioactive decay is an exponential decay, and is given as:
$N = N_0e^{-\lambda t}$, where $N_0$ is the initial quantity of the radioactive substance, N is the remaining quantity after time t, and $\lambda$ is the decay constant. A negative sign indicates that there is a decrease in the material remaining with time.
Now, the half-life of a radioactively decaying substance is defined as the time required for the substance to reduce to half its initial quantity. Therefore, if $t_{1/2}$ is the half life of a substance at which $N = \dfrac{N_0}{2}$, the equation becomes:
$\dfrac{N_0}{2} = N_0e^{-\lambda t_{1/2}} \Rightarrow e^{-\lambda t_{1/2}} = \dfrac{1}{2}$
Now, rewriting the first equation by multiplying and dividing the power of the exponential by $-\lambda t_{1/2}$:
$N = N_0e^{-\lambda t.\dfrac{-\lambda t_{1/2}}{-\lambda t_{1/2}}} = N_0e^{(-\lambda t_{1/2}).\dfrac{-\lambda t}{-\lambda t_{1/2}}}$
Now, substituting $ e^{-\lambda t_{1/2}} = \dfrac{1}{2}$ we get:
$N= N_0\left(\dfrac{1}{2}\right)^{\dfrac{-\lambda t}{-\lambda t_{1/2}}} \Rightarrow N = N_0\left(\dfrac{1}{2}\right)^{\dfrac{t}{t_{1/2}}}$
Now, we are given that for the radioactive carbon-14, $t_{1/2} = 5700\;yrs$ and $t = 17,100\;yrs$.
 Substituting this in the expression that we derived we get:
$N = N_0\left(\dfrac{1}{2}\right)^{\dfrac{17100}{5700}} = N_0 \left(\dfrac{1}{2}\right)^{3} = \dfrac{N_0}{2^3}$
$\Rightarrow N = \dfrac{1}{8}\times N_0$.
Therefore, there is one-eighth of the initial quantity left after a period of 17,100 years.

Note: Remember that there is an alternate form of expressing the exponential radioactive decay:
$N=N_0e^{\dfrac{-t}{\tau}}$, where $\tau$ is the mean lifetime of the decaying quantity.
Also, the decay constant $\lambda$ can be expressed in terms of half-life by taking $N = \dfrac{N_0}{2}$ at $t=t_{1/2}$:
$\dfrac{N_0}{2} = N_0 e^{-\lambda t_{1/2}} \Rightarrow e^{-\lambda t_{1/2}} = \dfrac{1}{2} \Rightarrow -\lambda t_{1/2} = ln\left(\dfrac{1}{2}\right) = ln\; 1 – ln\; 2 = 0 -0.693 = -0.693$
$\Rightarrow \lambda = \dfrac{0.693}{t_{1/2}}$