A) 1.78 per sec
B) 0.78 per sec
C) 2.78 per sec
D) 5.78 per sec
Correct Answer: D
Solution :
[d] at time \[t=0;\] \[{{A}_{0}}=\frac{d{{N}_{0}}}{dt}\] \[\frac{A}{{{A}_{0}}}=\frac{dN/dt}{d{{N}_{0}}/dt}=\frac{3260}{6520}\] or \[\lambda =\frac{2.303}{t}\log \frac{{{A}_{0}}}{A}=\frac{2.303}{2\times 60}\log 2\] \[=5.78\,\,per\,\,\sec \]
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