A piece of wood is found to have the `(C^(14))/(C^(12))` ratio to be `0.5` times of that in a living plant The number of years back the plant died will be ( `T` of `C^(14)=5,580` years)

A piece of wood was found to have C^(14)/C^(12) ratio 0.7 times that in a living plant.Calculate the period when the plant died. Half-life of C^(14)= 5760 years

A piece of wood was found to have C^(14)//C^(12) ratio 0.6 times that in a living plant. Calculate that in a living plant. Calculate the period when the plant died. (Half life of C^(14) = 5760 years)?

A piece of wood was found to have C^(14)//C^(12) ratio 0.6 times that in a living plant. Calculate that in a living plant. Calculate the period when the plant died. (Half life of C^(14) = 5760 years)?

A piece of charcoal from the ruins of a settlement in Japan was found to have .^(14)C// .^(14)C ratio what was 0.617 times that found in living organisms. How old is the piece of charcol? ( t_(1//2) for .^(14)C is 5770 year?)

The C-14 content of an ancient piece of wood was found to have three tenths of that in living trees. How old is that piece of wood? (log 3= 0.4771, log 7 = 0.8540 , Half-life of C-14 = 5730 years )

The amount of ._(6)^(14)C isotope in a piece of wood is found to one fourth (1/4) of that present in a fresh piece of wood. Calculate the age of the piece of wood (t_(12) "of "_(6)^(14)C=5770 "years") a) 7999 year b) 11543 year c) 16320 year d) 23080 year