A) 40 year, 0.9173/year
B) 90 year, 9.017/year
C) 80 year, 0.0173 year
D) None of these
Correct Answer: C
Solution :
To reduce one fourth it takes time \[t=2\left( {{T}_{1/2}} \right)\]= 2 ´ 40 = 80 years. Decay constant \[\lambda =\frac{0.693}{{{T}_{1/2}}}=\frac{0.693}{40}=0.0173\,years\]You need to login to perform this action.
You will be redirected in
3 sec