A) 3.2 Tyr
B) 4.6 Tyr
C) 6.6 Tyr
D) 9.2 Tyr
Correct Answer: C
Solution :
From the question,N \[=\frac{{{N}_{0}}}{100}\] and \[{{t}_{1/2}}=T\,\,yr\] So, \[\lambda =\frac{0.693}{T}\] Thus, \[t=\frac{2.303}{\lambda }\log \left[ \frac{{{N}_{0}}}{N} \right]=\frac{2.303T}{0.693}\log \,100\] = 6.65T
You need to login to perform this action.
You will be redirected in
3 sec
