A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ___________. (Nearest integer)
(Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)

Solution

Chemistry ma’am · 4 steps

Get k from the half-life, put it into the log form of first-order decay, and the given antilog finishes it.

1. Rate constant. Radioactive decay is first order, so k = 0.693/t_(1/2) = 0.693/200 day⁻¹. Activity is proportional to the number of nuclei left, so A₀/A = N₀/N.

   log A₀/A = (k t)/2.303 = (0.693 × 83)/(200 × 2.303)

2. Put in numbers. The numerator is 0.693 × 83 = 57.52 and the denominator is 200 × 2.303 = 460.6. So log A₀/A = 0.1249 ≈ 0.125.
3. Take the antilog. A₀/A = antilog 0.125 = 1.333, so A/A₀ = 1/1.333 = 0.75.
4. So. The activity remaining is 0.75 × 100 = 75% of the original.

The common slip: 25. Reporting the fraction that has decayed instead of the fraction that remains gives 25, not 75.

The idea to keep. For any decay question, write k = 0.693/t_(1/2) first, then use log(N₀/N) = kt/2.303. Check which way up the ratio is before you take the antilog.

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