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

Solution

Physics sir · 4 steps

One-eighth is three halvings in three days, so the half-life is one day. Now run time backwards.

1. Find the half-life. 1/8 = (1/2)³, so the material went through 3 half-lives in 3 days. The half-life is T_(1/2) = 1 day.
2. Count half-lives. In 5 days there are 5/1 = 5 half-lives. The fraction left is (1/2)⁵ = 1/32.

   N = N₀(1/2)^(t/T_(1/2)) ⇒ N₀ = 32 N

3. Put in numbers. The amount left is 8 × 10⁻³ kg = 8 g. So N₀ = 32 × 8 g = 256 g.
4. So. The initial amount was 256 g, option (b).

The idea to keep. Write any fraction left as a power of 1/2. The power is the number of half-lives, and that gives you the half-life straight away.

Answer from the NTA final answer key. Solution by PrepWiser.