A) \[3.1\times {{10}^{-8}}g\]
B) \[1.55\times {{10}^{-8}}\,g\]
C) \[6.2\times {{10}^{-8}}g\]
D) \[3.1\times {{10}^{-10}}g\]
Correct Answer: A
Solution :
We know that, \[\frac{-dN}{dt}=\lambda .\frac{6.023\times {{10}^{23}}\times \omega }{M}\] Here, \[\frac{-dN}{dt}=1\,\text{curie}\] \[=3.7\times {{10}^{10}}\,\text{dps}\] \[\lambda =\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{26.8\times 60}{{\sec }^{-1}}\] \[\omega =?,\,\text{M=214}\] Hence, \[3.7\times {{10}^{10}}=\frac{0.693}{26.8\times 60}\times \frac{6.023\times {{10}^{23}}\times \omega }{214}\] \[\therefore \] \[\omega =3.1\times {{10}^{-8}}\,g\]
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