question_answer

The half life for a-decay of uranium \[_{\text{92}}{{\text{U}}^{\text{228}}}\] is \[4.47\times {{10}^{8}}\] yr. If a rock contains 60% of      original \[_{\text{92}}{{\text{U}}^{\text{228}}}\] atoms, then its age is [take log 6 =0.778, log 2=0.3]

A) \[1.2\times {{10}^{7}}yr\]            

B) \[3.3\times {{10}^{8}}yr\]

C) \[4.2\times {{10}^{9}}yr\]            

D) \[6.5\times {{10}^{9}}yr\] Given \[\left( \frac{2E}{{{r}_{1}}+{{r}_{2}}+R} \right){{r}^{2}}\]               \[E-\frac{2E}{\left( {{r}_{1}}+{{r}_{2}}+R \right)}{{r}_{2}}=0\] \[{{r}_{1}}+{{r}_{2}}+R=2{{r}_{2}}\]            \[R={{r}_{2}}-{{r}_{1}}\] Apply logarithm on both sides           \[~n\log 2=\log 10log6\] \[\therefore \]     \[Bqv=\frac{m{{v}^{2}}}{r}\]            \[{{m}_{e}}{{v}_{e}}={{m}_{p}}{{v}_{p}}\] So     \[{{v}_{p}}=\left( \frac{{{m}_{e}}}{{{m}_{p}}} \right){{v}_{e}}=\left( \frac{9\times {{10}^{-31}}}{1.8\times {{10}^{-27}}} \right)3\times {{10}^{6}}\]           \[{{v}_{p}}=1.5\times {{10}^{3}}m/s\]