A) 0.5
B) 0.25
C) 0.125
D) 0.33
Correct Answer: C
Solution :
If N be the number of atoms of a radioactive substance left at some instant of time, than \[\Rightarrow \] where \[{{N}_{0}}\] is the initial number of atoms and n is number of half lives. \[\int\limits_{x=20}^{x=30}{(-0.1)x\,dx}\,={{K}_{f}}-500\] \[\int{{{x}^{n}}dx}=\frac{{{x}^{n+1}}}{n+1},\] \[-0.1\left[ \frac{{{x}^{2}}}{2} \right]_{x=20}^{x=30}={{K}_{f}}-500\] \[-0.1\left[ \frac{{{(30)}^{2}}}{2}-\frac{{{(20)}^{2}}}{2} \right]={{K}_{f}}-500\] \[\Rightarrow \]You need to login to perform this action.
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