A) \[{{O}^{8}}\]
B) \[P{{b}^{212}}\]
C) \[F{{e}^{59}}\]
D) \[_{6}{{C}^{14}}\]
Correct Answer: D
Solution :
The isotopes \[_{6}{{C}^{14}}\] is radioactive. Half-life of \[_{6}{{C}^{14}}\] is \[5770\text{ }yr\]. \[_{6}^{14}C\] isotope produced in upper atmosphere is incorporated in \[C{{O}_{2}}\] which is inhaled by plants and in turn consumed by animals and human beings \[_{7}^{14}N+_{0}^{1}n\xrightarrow{{}}_{6}^{16}C+_{1}^{1}H\] A plant-animal carbon cycle exists in nanas and as long as it is alive, the \[_{6}^{14}C\] content remains constant when a plant or animal dies, the process of intake of \[C-14\] stops and it stars decaying \[_{6}^{14}C\xrightarrow{{}}_{7}^{14}N+_{-1}^{e}e;\] \[{{t}_{1/2}}=5770yr\] By knowing the equilibrium concentration of \[_{6}^{14}C\]in living plants and \[_{6}^{14}C\] in a dead plant, the age can be calculated by using the following expression. \[t=\frac{2.303}{\lambda }\log \frac{{{N}_{0}}}{{{N}_{t}}}\] \[\left( \lambda =\frac{0.63}{5770}y{{r}^{-1}} \right)\]
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