A) \[2.95\times {{10}^{11}}yr\]
B) \[2.95\times {{10}^{9}}yr\]
C) \[4.37\times {{10}^{9}}yr\]
D) \[4.37\times {{10}^{11}}yr\]
Correct Answer: C
Solution :
[c] If \[{{N}_{0}}\] potassium atoms were present at the time the rock was formed by solidification from a molten form, the number of potassium atoms remaining at the time of analysis is \[{{N}_{K}}={{N}_{0}}{{e}^{-\lambda t}}\] ...(i) in which t is the age of the rock. For every potassium atom that decays, an argon atom is produced. Thus, the number of argon atoms present at the line of the analysis is \[{{N}_{Ar}}={{N}_{0}}-{{N}_{K}}\] ...(it) We cannot measure No, so let's eliminate it from eqs. and (ii). We find, after some algebra, that \[\lambda t=\ell n\left( 1+\frac{{{N}_{Ar}}}{{{N}_{K}}} \right),\] in which \[{{N}_{Ar}}/{{N}_{K}}\] can be measured. Solving for t, \[t=\frac{{{T}_{1/2}}In\,(1+{{N}_{A}}/{{N}_{K}})}{In\,\,2}\] \[=\frac{(1.25\times {{10}^{9}}y)\,[In\,(1+10.3)]}{In\,\,2}=4.37\times {{10}^{9}}yr\]
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