Question

The fraction of a radioactive material that remains active after time $t$ is $\frac{9}{16}$. The fraction which remains active after time $\frac{t}{2}$ will be

• A. $\frac{4}{5}$
• B. $\frac{7}{8}$
• C. $\frac{3}{5}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D

$\frac{3}{4}$

Step 1: Given

Radioactive material active after time $t$: $\frac{N\left(t\right)}{N_0}=\frac{9}{16}$

Radioactive material active after time $\frac{t}{2}$: $\frac{N\left(\frac{t}{2}\right)}{N_0}=?$

Step 2: Formula used

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

Where $N\left(t\right)$ is the concentration at time $t$, $N_0$ is the initial concentration, $\lambda$ is the rate of decay.

Step 3: Determine the value of $e^{-\lambda t}$

Using the equation for first-order decay,

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

Where $N\left(t\right)$ is the concentration at time $t$, $N_0$ is the initial concentration, $\lambda$ is the rate of decay.

$$\begin{gathered} \frac{N\left(t\right)}{N_0}=e^{-\lambda t} \\ \frac{9}{16}=e^{-\lambda t}-\left(1\right) \end{gathered}$$

Step 4: Determine the value of $N\left(\frac{t}{2}\right)$

Using the equation for first-order decay again,

$$\begin{gathered} N\left(\frac{t}{2}\right)=N_0{e}^{-\lambda \frac{t}{2}} \\ \frac{N\left(\frac{t}{2}\right)}{N_0}=e^{-\lambda \frac{t}{2}} \\ =\sqrt{e^{-\lambda t}} \end{gathered}$$

Substituting equation (1)

$$\begin{gathered} \frac{N\left({\displaystyle \frac{t}{2}}\right)}{N_0}=\sqrt{e^{-\lambda t}}=\sqrt{\frac{9}{16}} \\ =\frac{3}{4} \\ N\left(\frac{t}{2}\right)=\frac{3}{4}{N}_0 \end{gathered}$$

Therefore, option D is the correct option.