A) 12.5 g
B) 10.5 g
C) 6.03 g
D) 4.03 g
Correct Answer: C
Solution :
Remaining material \[N=\frac{{{N}_{0}}}{{{2}^{t/T}}}\] \[\Rightarrow N=\frac{10}{{{(2)}^{20/15}}}=\frac{10}{2.15}=3.96\ gm\] So decayed material \[=10-3.96=6.04\ gm\]You need to login to perform this action.
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