A) 100 years
B) 50 years
C) 5 years
D) 10 years
Correct Answer: D
Solution :
\[K=\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{10\,\text{years}}\] If initial concentration \[a=10\ gm\]and final concentration \[x=\frac{a}{2}=5\ gm\] then, \[t=\frac{2.303}{K}\log \frac{a}{a-x}\] \[=\frac{2.303}{.693}\times 10\times \log \frac{10}{5}\] \[=\frac{2.303\times 10\times \log 2}{.693}\]\[=\frac{2.303\times 10\times 0.301}{0.693}=10\ \text{years}\].
You need to login to perform this action.
You will be redirected in
3 sec
