The half-life of radioisotope bromine - 82 is 36 hours. The fraction which remains after one day is ________ × 10⁻².
(Given antilog 0.2006=1.587)

Solution

Chemistry ma’am · 4 steps

One day is two-thirds of a half-life, so the fraction left is (1/2)^(2/3). Logs do the rest.

1. Count half-lives. One day is 24 hours and the half-life is 36 hours. So the number of half-lives is n = 24/36 = 2/3. It does not need to be a whole number.

   N/N₀ = (1/2)ⁿ = (1/2)^(2/3) = 1/2^(2/3)

2. Use logs. log 2^(2/3) = 2/3 × 0.3010 = 0.2007. The given antilog of 0.2006 is 1.587, so 2^(2/3) = 1.587.
3. Take the reciprocal. N/N₀ = 1/1.587 = 0.630, which is 63 × 10⁻².
4. So. The fraction of bromine-82 that remains after one day is 63 × 10⁻².

The common slip: 37. Calculating the fraction that has decayed, 1 - 0.63, gives 37. The question asks for the fraction that remains.

The idea to keep. Radioactive decay is first order, so the fraction left is (1/2)^(t/t_(1/2)) even when t/t_(1/2) is a fraction. Take logs to handle odd powers like 2/3.

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