Nuclear Chemistry Question 23
Question: The activity of carbon-14 in a piece of an ancient wood is only 12.5%. If the half-life period of carbon-14 is 5760 years, the age of the piece of wood will be $ (\log 2=0.3010) $
[MP PMT 1999]
Options:
A) $ 17.281\times 10^{2} $ years
B) $ 172.81\times 10^{2} $ years
C) $ 1.7281\times 10^{2} $ years
D) $ 1728.1\times 10^{2} $ years
Show Answer
Answer:
Correct Answer: B
Solution:
$ {t _{1/2}} $ of C-14 = 5760 year, $ \lambda =\frac{0.693}{5760}, $ Now $ t=\frac{2.303}{\lambda }\log \frac{^{14}Coriginal}{^{14}Caftertimet} $
$ =\frac{2.303\times 5760}{0.693}\log \frac{100}{12.5}=\frac{2.303\times 5760\times 0.9030}{0.693} $ = 17281= 172.81 × $ 10^{2} $ years.
BETA
