A) 1
B) 3
C) 5
D) 6
Correct Answer: A
Solution :
We know that, \[\left| \frac{dN}{dt} \right|=\lambda N=\frac{1}{{{T}_{mean}}}N\] \[\therefore \,\,\,{{10}^{10}}=\frac{1}{{{10}^{9}}}\times N\] \[\therefore \,\,N={{10}^{19}}\] i.e., \[{{10}^{19}}\] radioactive atoms are present in the freshly prepared sample. The mass of the sample \[\therefore \,\,={{10}^{19}}\times {{10}^{-25}}kg={{10}^{-6}}kg=1mg\]You need to login to perform this action.
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