A) \[1.64\times {{10}^{-10}}yrs\]
B) \[7.14\times {{10}^{9}}yrs\]
C) \[2.25\times {{10}^{9}}yrs\]
D) None of these
Correct Answer: B
Solution :
\[t=\frac{1}{\lambda }\] In \[\frac{{{n}_{U}}+{{n}_{Pb}}}{{{n}_{U}}};{{n}_{U}}=\]no. of mole of \[{{U}^{238}}\] \[t=\frac{1}{\lambda }\] In \[\left( \frac{0.15}{0.05} \right);\] \[{{n}_{Pb}}=\]no-of mole of \[P{{b}^{206}}\] \[t=\frac{1}{\lambda }\] In \[(3)=\frac{1.1}{0.693}\times 4.5\times {{10}^{9}}=7.14\times {{10}^{9}}\]years
You need to login to perform this action.
You will be redirected in
3 sec
