Question

Find the half life of a radioactive element, if its activity decreases for $1$ month by $10\%$.

• A. $193.3$ days
• B. $197.3$ days
• C. $198.5$ days
• D. $199.7$ days

Answer 1

Answer: (B)

$197.3$ days

Activity of an isotope is measured by the number of nuclei decaying for a time unit.

Suppose that $d{N}_d$ nuclei decay for a short period of time $dt$.

Then the isotope activity $A$ is expressed by the formula $A=\frac{d{N}_d}{dt}$

It follows from the radioactive decay law that

$N\left(t\right)=N_0{e}^{-\lambda t}$ where $N\left(t\right)$ is the quantity of the remaining nuclei.

Therefore,

$N_d\left(t\right)=N_0-N\left(t\right)=N_0-N_0{e}^{-\lambda t}=N_0(1-e^{-\lambda t})$

By differentiating with respect to $t$, we find the expression for activity:

$A\left(t\right)=\frac{d{N}_d}{dt}=N_0\lambda e^{-\lambda t}$

The initial isotope activity is equal to

$A(t=0)=A_0=N_0\lambda$

Hence, $A\left(t\right)=A_0{e}^{-\lambda t}$

As it can be seen, the activity decreases over time by the same law as the amount of undecayed material. Substituting the expression for the half life $T=\frac{ln2}{\lambda }$ in the last formula, we can write:

$A\left(t\right)=A_0{e}^{-\frac{tln2}{T}}$

The value of $T$ in the last expression can be found by

$e^{-\frac{tln2}{T}}=\frac{A}{A_0}$

$\Rightarrow -\frac{tln2}{T}=ln\frac{A}{A_0}$

$\Rightarrow \frac{tln2}{T}=ln\frac{A_0}{A}$

$\Rightarrow T=\frac{tln2}{ln\frac{A_0}{A}}$

In our case, the half life period of the given isotope is

$T=\frac{tln2}{ln\frac{A_0}{A}}=\frac{30ln2}{ln\frac{100}{90}}\approx \frac{30\cdot 0.93}{ln1.11}\approx 197.3days$