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Exercises · 8.10

Q.Show that the mass of radium (88226Ra^{226}_{88}\text{Ra}) with an activity of 1 curie is almost a gram. Given T1/2=1600T_{1/2}=1600 years.

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Step 1. The decay constant of radium-226 is λ=ln⁡2T1/2=0.69311600×3.156×107 s≈1.37×10−11 s−1\lambda=\dfrac{\ln2}{T_{1/2}}=\dfrac{0.6931}{1600\times3.156\times10^7\ \text{s}}\approx1.37\times10^{-11}\ \text{s}^{-1} (using 11 year =3.156×107=3.156\times10^7 s).

Step 2. For an activity of exactly 11 curie, R=3.7×1010R=3.7\times10^{10} Bq (decays/s). From R=λNR=\lambda N, the required number of radium-226 atoms is

N=Rλ=3.7×10101.37×10−11≈2.70×1021 atomsN=\frac{R}{\lambda}=\frac{3.7\times10^{10}}{1.37\times10^{-11}}\approx2.70\times10^{21}\ \text{atoms}

Step 3. Converting this atom count to a mass using Avogadro's number (6.02×10236.02\times10^{23} atoms per 226 g of radium-226, since its atomic mass is 226 u):

m=N6.02×1023×226 g=2.70×10216.02×1023×226≈4.49×10−3×226≈1.01 gm=\frac{N}{6.02\times10^{23}}\times226\ \text{g}=\frac{2.70\times10^{21}}{6.02\times10^{23}}\times226\approx4.49\times10^{-3}\times226\approx1.01\ \text{g} …

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