A) \[45.3\times {{10}^{9}}\] years
B) \[45.3\times {{10}^{10}}\] years
C) \[4.53\times {{10}^{9}}\] years
D) \[4.53\times {{10}^{10}}\] years
Correct Answer: C
Solution :
According to radioactive equilibrium \[{{\lambda }_{A}}{{N}_{A}}={{\lambda }_{B}}{{N}_{B}}\] or \[\frac{0.693\times {{N}_{A}}}{{{t}_{1/2}}(A)}=\frac{0.693\times {{N}_{B}}}{{{t}_{1/2}}\,(B)}\left[ \lambda =\frac{0.693}{{{t}_{1/2}}} \right]\] Where \[{{t}_{1/2}}(A)\] and \[{{t}_{1/2}}(B)\] are half periods of A and B respectively \[\therefore \frac{{{N}_{A}}}{{{t}_{1/2}}(A)}=\frac{{{N}_{B}}}{{{t}_{1/2}}(B)}\,\,\text{or}\,\,\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{{{t}_{1/2}}(A)}{{{t}_{1/2}}(B)}\] \[\therefore \] At equilibrium A and B are present in the ratio of their half lives \[\frac{1}{2.8\times {{10}^{6}}}=\frac{1620}{\text{Half}\,\text{life}\,\,\text{of}\,\text{uranium}}\] \[\therefore \]Half-life of uranium = \[2.8\times {{10}^{6}}\times 1620=4.53\times {{10}^{9}}\]years.You need to login to perform this action.
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