Question

In a radioactive material, the fraction of an active material remaining after time $t$ is $\frac{9}{16}$. The fraction that was remaining after $\frac{t}{2}$ is-

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

Answer 1

The correct option is C $\frac{3}{4}$

As we know, for first order decay,

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

According to equation,

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

After time, $\frac{t}{2}$

$N\left(t/2\right)=N_0{e}^{-\lambda t}$

$\frac{N\left(t/2\right)}{N_0}=\sqrt{e^{-\lambda t}}=\sqrt{\frac{9}{16}}$

$\therefore N\left(t/2\right)=\frac{3}{4}{N}_0$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.