Half-life problems — if 100g of radioactive substance has half life 10 years find mass after 30 years

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Question

A radioactive substance has a half-life of 10 years. If we start with 100 g of this substance, how much will remain after 30 years?

Solution — Step by Step

The half-life (t1/2t_{1/2}) is the time required for exactly half of the radioactive nuclei in a sample to decay. After every half-life, the amount is reduced by half.

This is a fixed property of the substance — it doesn’t matter how much you start with. 100 g becomes 50 g in 10 years; 1 kg becomes 500 g in 10 years.

Total time = 30 years. Half-life = 10 years.

Number of half-lives = 3010=3\dfrac{30}{10} = 3

Starting with 100 g:

Time (years)Half-lives elapsedAmount remaining
00100 g
10150 g
20225 g
30312.5 g
m=m0×(12)nm = m_0 \times \left(\frac{1}{2}\right)^n

where nn = number of half-lives = 3.

m=100×(12)3=100×18=12.5 gm = 100 \times \left(\frac{1}{2}\right)^3 = 100 \times \frac{1}{8} = 12.5 \text{ g}

Why This Works

Radioactive decay is a first-order process — the rate of decay is proportional to the amount of substance present. This mathematical property produces the exponential decay curve, and the half-life is constant regardless of how much material is present.

The general formula is:

m=m0eλt=m0(12)t/t1/2m = m_0 \cdot e^{-\lambda t} = m_0 \cdot \left(\frac{1}{2}\right)^{t/t_{1/2}}

where λ=ln2t1/20.693t1/2\lambda = \dfrac{\ln 2}{t_{1/2}} \approx \dfrac{0.693}{t_{1/2}} is the decay constant.

Alternative Method — Using Decay Constant

λ=0.693t1/2=0.69310=0.0693 yr1\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{10} = 0.0693 \text{ yr}^{-1} m=100×e0.0693×30=100×e2.079m = 100 \times e^{-0.0693 \times 30} = 100 \times e^{-2.079} e2.079=eln8=18=0.125e^{-2.079} = e^{-\ln 8} = \frac{1}{8} = 0.125 m=100×0.125=12.5 gm = 100 \times 0.125 = 12.5 \text{ g}

Same answer, as expected.

Common Mistake

Students sometimes think “50% decays per half-life, so 3 × 50% = 150% decays in 3 half-lives.” Percentages don’t add like that. Each half-life halves the remaining amount, not the original. After 3 half-lives, 12×12×12=18\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} of the original remains — that’s 12.5%, not 50%.

A quick table to memorise: after nn half-lives, fraction remaining = (1/2)n(1/2)^n. So: 1 → 1/2, 2 → 1/4, 3 → 1/8, 4 → 1/16, 5 → 1/32, 10 → 1/1024. After 10 half-lives, less than 0.1% of the original remains.

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