Question

A radio-isotope has a half-life of 5 years. The fraction of the atoms of this material that would decay in 15 years will be
A. $\dfrac{1}{8} \\ $
B. $\dfrac{2}{3} \\ $
C. $\dfrac{7}{8} \\ $
D. $\dfrac{5}{8}$

Answer 1

Hint: First determine the value of decay constant using the half life equation. Then, find the value of $\dfrac{N}{N_{0}}$ by using the exponential decay equation. After that, subtract the obtained value from 1 because it is mentioned we need to find the fraction of the atoms of the given material that would decay in 15 years.

Formula used:
$N=N_{0}e^{-\lambda~t}$
$T_{1/2}=\dfrac{\log_{e}{2}}{\lambda}$
Where, $T_{1/2}$ is the half life of the radioactive substance, $\lambda$ is the decay constant.

Complete step by step solution:
As mentioned in the question the half-life of radio-isotope is 5 years. The mathematical form of exponential radioactive law can be written as follows:
$N=N_{0}e^{-\lambda~t}$
But we need to determine the fraction of the atoms of the given material that would decay in 15 years. As a result, the above equation can be written as follows:
$\dfrac{N}{N_{0}}=e^{-\lambda~t}$ ......(1)
Also, one of the equations for the half-life is,
$T_{1/2}=\dfrac{\log_{e}{2}}{\lambda} \\ $ ......(2)
It is given that $T_{1/2}=5~years$ ........(3)

By equating the equations (2) and (3), we get
$T_{1/2}=\dfrac{\log_{e}{2}}{\lambda}=5~years$
$\Rightarrow\lambda=\dfrac{\log_{e}{2}}{5} \\ $
Also, t=15 years (given)
Let's substitute the values of t and in equation (1)
$\dfrac{N}{N_{0}}=e^{-\dfrac{\log_{e}{2}}{5}\times15} \\ $
$\Rightarrow\dfrac{N}{N_{0}}=e^{-3{\log_{e}{2}}}$

For solving RHS, let's consider RHS=y, i.e,
$e^{-3{\log_{e}{2}}}=y \\ $
Taking $\log_{e}$ on both sides,
$\log_{e}e^{-3{\log_{e}{2}}}=\log_{e}{y} \\ $
The power of e will then appear in the front.
$-3\log_{e}{2}\log_{e}{e}=\log_{e}{y} \\ $
We know $\log_{e}{e}=1$
Now,
$-3\log_{e}{2}=\log_{e}{y} \\ $
Take -3 in the power of 2.
$\log_{e}{2^{-3}}=\log_{e}{y}$

Canceling log on both sides, we get
$2^{-3}=y$
This implies,
$\lgroup\dfrac{1}{2}\rgroup^{3}=y$
Therefore,
$y=\dfrac{1}{8}$
So, $\dfrac{N}{N_{0}}=\dfrac{1}{8}$
Therefore, a decaying fraction will be
$1-\dfrac{1}{8}=\dfrac{7}{8}$

Hence, the correct option is C.

Note: There is also an alternate way to solve this question by using the equation of exponential decay law independent of the decay constant. The equation is $N=N_{0}\lgroup\dfrac{1}{2}\rgroup^{\dfrac{t}{T}}$. By using this equation we can get $\dfrac{N}{N_{0}}$ easily.