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Exercise 5 · Q7

Q.Radium decomposes at a rate proportional to the amount present. If half the original amount disappears in 1600 years, find the percentage lost in 100 years.

Lakshadweep CbseNCERTSubjective· 3mImportance★★★★★
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With decay constant k=log⁡21600k=\frac{\log 2}{1600}, the fraction left after 100 years is e−100k=0.9576e^{-100k}=0.9576, so about 4.24% is lost.

N(t)=N0e−ktN(t)=N_0e^{-kt} — radioactive decay, where N0N_0 = original amount, N(t)N(t) = amount left after time tt, kk = decay constant, tt = time in years. Half-life relation: k=log⁡2t1/2k=\dfrac{\log 2}{t_{1/2}}.

  1. Find kk from the half-life. Half disappears in 1600 years, so 12N0=N0e−1600k⇒k=log⁡21600=0.69311600=4.3322×10−4 yr−1\dfrac12 N_0=N_0e^{-1600k}\Rightarrow k=\dfrac{\log 2}{1600}=\dfrac{0.6931}{1600}=4.3322\times10^{-4}\ \text{yr}^{-1}.
  2. Amount left after 100 years. NN0=e−100k=e−100×4.3322×10−4=e−0.043322\dfrac{N}{N_0}=e^{-100k}=e^{-100\times 4.3322\times10^{-4}}=e^{-0.043322}. …

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