Question

The half-life of a radioactive substance is 20 minutes. $\frac{1}{3}$rd part of substance has decayed in time $t_1$ and $\frac{2}{3}$rd part of it has decayed in time $t_2$. Then, ($t_2 - t_1$) is nearly

• A. 7 minutes
• B. 14 minutes
• C. 20 minutes
• D. 28 minutes

Hint:

For any first-order process, the time taken to decay from fraction $f_1$ to $f_2$ depends only on the ratio of the remaining quantities.
Since the ratio of the remaining amounts $\frac{N(t_1)}{N(t_2)} = \frac{2/3}{1/3} = 2$, this decay corresponds exactly to one half-life.

Answer 1

The Correct Option is C

Step 1: Express both times using half-lives directly, in base 2.
Fraction remaining is $N/N_0 = (1/2)^{t/T}$, so $t/T = \log_2(N_0/N)$, where $T = 20$ minutes is the half-life.
Step 2: Write $t_1$ for one-third decayed (two-thirds remaining). \[ \frac{t_1}{T} = \log_2\left(\frac{3}{2}\right) \]
Step 3: Write $t_2$ for two-thirds decayed (one-third remaining). \[ \frac{t_2}{T} = \log_2(3) \]
Step 4: Subtract, using log rules to combine them into one clean log. \[ \frac{t_2-t_1}{T} = \log_2(3) - \log_2\left(\frac{3}{2}\right) = \log_2\left(\frac{3}{3/2}\right) = \log_2(2) = 1 \] \[ \boxed{t_2 - t_1 = T = 20\text{ minutes}} \]