Question

The half-life of Thorium-227 is 18.72 days. How many days are required for 75% of a given amount to decay?

Answer 1

Hint: Recall that the quantity remaining at half-life is half the initial quantity of the radioactive substance. Use the exponential decay equation to determine the relation between the amount decayed, time taken and the half-life period, and proportionally evaluate the relation to arrive at the time taken until there is only twenty-five percent of the radioactive substance left. 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 answer:
Let $N = N_0$ be the initial amount of Thorium-227 at $t =0$.
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. 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.
We are given that the half-life of Thorium-227 is 18.72 days, i.e.,
$N = \dfrac{N_0}{2}$ at $t = t_{1/2} = 18.72\;days$
Plugging this into the radioactive day equation we get:
$\dfrac{N_0}{2} = N_0e^{-\lambda t_{1/2}} \Rightarrow e^{-\lambda t_{1/2}} = \dfrac{1}{2}$
Now, rewriting the decay 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:
$N= N_0\left(\dfrac{1}{2}\right)^{\dfrac{-\lambda t}{-\lambda t_{1/2}}} \Rightarrow N = N_0\left(\dfrac{1}{2}\right)^{\dfrac{t}{t_{1/2}}}$
Let the time taken to decay $75\%$ of it be$\;t$. If $75\%$ of the initial amount has decayed, it means that $25\%$ of the initial amount is left, i.e.,
$N = \dfrac{25}{100} \times N_0 = \dfrac{N_0}{4}$
Plugging this into the expression we derived, we get:
$\dfrac{N_0}{4} = N_0\left(\dfrac{1}{2}\right)^{\dfrac{t}{t_{1/2}}}$
$\Rightarrow \dfrac{1}{4} = \left(\dfrac{1}{2}\right)^{\dfrac{t}{t_{1/2}}}$
$\Rightarrow \left(\dfrac{1}{2}\right)^2 = \left(\dfrac{1}{2}\right)^{\dfrac{t}{t_{1/2}}}$
Equating the powers we get:
$2 = \dfrac{t}{t_{1/2}} \Rightarrow t = 2 \times t_{1/2}$
Substituting $t_{1/2} = 18.72\; days$, we get:
$t = 2 \times 18.72 = 37.44\;days$
Therefore, it takes $37.44\;days$ for $75\%$ of the Thorium-227 sample to decay.

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 $t=t_{1/2}$:
$\dfrac{N_0}{2} = N_0 e^{-\lambda t_{1/2}} \Rightarrow e^{-\lambda t_{1/2}} = \dfrac{1}{2} \Rightarrow -\lambda t_{1/2} = ln\left(\dfrac{1}{2}\right) = ln\; 1 – ln\; 2 = 0 -0.693 = -0.693$
$\Rightarrow \lambda = \dfrac{0.693}{t_{1/2}}$