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 \[{{v}_{p}}=\left( \frac{{{m}_{e}}}{{{m}_{p}}} \right){{v}_{e}}=\left( \frac{9\times {{10}^{-31}}}{1.8\times {{10}^{-27}}} \right)3\times {{10}^{6}}\] \[{{v}_{p}}=1.5\times {{10}^{3}}m/s\]
Correct Answer: B
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