Assuming 1 μ g of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ______ × 10⁻¹ μ g.
[Given : ln 10 = 2.303; log 2 = 0.30]

Solution

Chemistry ma’am · 4 steps

Count the half-lives as 100/30, then use the given log 2 so the ratio comes out to exactly 10.

1. First-order decay. Radioactive decay is first order, so after n half-lives the amount left is N = N₀ (1/2)ⁿ. Here n = t/t_(1/2) = 100/30 = 10/3, which is not a whole number.
2. Take logs. Because n is a fraction, take log of both sides. This turns the power into a product.

   log N₀/N = n log 2 = 10/3 × 0.30 = 1

3. Find N. log N₀/N = 1 means N₀/N = 10. With N₀ = 1 μg, N = 0.1 μg = 1 × 10⁻¹ μg. The ln route gives the same: k = (2.303 × 0.30)/30, so kt = 2.303 = ln 10.
4. So. The amount of X left after 100 years is 1 × 10⁻¹ μg, so the answer is 1.

The common slip: 0.1. Writing the mass itself, 0.1, in the blank gives the wrong entry. The blank asks for the number that multiplies 10⁻¹, which is 1.

The idea to keep. When t is not a whole number of half-lives, use log N₀/N = t/t_(1/2) log 2. Use the log values given in the question, because they are chosen to make the numbers come out neatly.

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