Question

A radio active material is reduced to 1/8 of its original amount in 3 days.
If 8×10^-3 kg of the material is left after 5 days the initial amount of the material is

• A. 32g
• B. 40g
• C. 256g
• D. 64g

Answer 1

The Correct Option is C

To solve this problem, we will use the concept of radioactive decay, which is governed by the half-life formula. Radioactive decay is an exponential decay process described by the equation:

N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{\frac{1}{2}}}}

Where:

• N(t) is the amount of substance remaining at time t.
• N_0 is the initial amount of substance.
• t_{\frac{1}{2}} is the half-life of the substance.
• t is the elapsed time.

According to the problem, the substance is reduced to \frac{1}{8} of its original amount in 3 days. Since reducing to \frac{1}{8} means three half-lives have passed (as (\frac{1}{2})^3 = \frac{1}{8}), the half-life t_{\frac{1}{2}} can be calculated as:

3 \text{ days} = 3 \times t_{\frac{1}{2}}

Thus, t_{\frac{1}{2}} = 1 \text{ day}.

Now, if 8 × 10^-3 kg (which is 8 g since 1 kg = 1000 g) is left after 5 days, we can use the same formula to find the initial quantity N_0:

8 = N_0 \times \left(\frac{1}{2}\right)^{\frac{5}{1}}

Simplifying this gives:

8 = N_0 \times \left(\frac{1}{2}\right)^5

8 = N_0 \times \frac{1}{32}

Solving for N_0, we get:

N_0 = 8 \times 32

N_0 = 256 \text{ g}

Therefore, the initial amount of the radioactive material was 256 g. This matches the correct answer:

• 32g
• 40g
• 256g (Correct Answer)
• 64g