A) \[1.7\times {{10}^{9}}\]
B) \[1.7\times {{10}^{11}}\]
C) \[2.9\times {{10}^{11}}\]
D) \[3.4\times {{10}^{9}}\]
Correct Answer: B
Solution :
No. of \[\alpha -\]particles per second \[=A=-\frac{dN}{dt}=N\lambda \]is required where \[N=\frac{6.0\times {{10}^{-23}}}{210}\times 1\times {{10}^{-3}}\]and \[\lambda =5.8\times {{10}^{-8}}\]per second So, \[A=N\lambda =\frac{6.0\times {{10}^{-23}}}{210}\times 1\times {{10}^{-3}}\times 5.8\times {{10}^{-8}}\] \[\approx 1.7\times {{10}^{11}}\] Hence, the correction option is (b).
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