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A piece of wood from the ruins of an anc...

A piece of wood from the ruins of an ancient building was found to have a `.^(14)C` activity of 12 disintergrations per minute per gram of its carbon content. The `.^(14)C` activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree from which the wooden sample came die ? Given halg-life of `.^(14)C` is 5760 yr.

A

1840

B

1830

C

1820

D

1860

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The correct Answer is:
A
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A piece of wood form the ruins of an ancient building was found to have a C^(14) activity of 12 disintegrations per minute per gram of its carbon content. The C^(14) activity of the living wood is 16 disintegrations/minute/gram. How long ago did the trees, from which the wooden sample came, die? Given half-life of C^(14) is 5760 years.

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Carbon -14 used to determine the age of organic material. The procedure is absed on the formation of C^(14) by neutron capture iin the upper atmosphere. ._(7)N^(14)+._(0)n^(1) rarr ._(6)C^(14)+._(1)H^(1) C^(14) is absorbed by living organisms during photosynthesis. The C^(14) content is constant in living organism. Once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of C^(14) in the dead being falls due to the decay, which C^(14) undergoes. ._(6)C^(14)rarr ._(7)N^(14)+beta^(c-) The half - life period of C^(14) is 5770 year. The decay constant (lambda) can be calculated by using the following formuls : lambda=(0.693)/(t_(1//2)) The comparison of the beta^(c-) activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method, however, ceases to be accurate over periods longer than 30000 years. The proportion of C^(14) to C^(12) in living matter is 1:10^(12) . A nuclear explosion has taken place leading to an increase in the concentration of C^(14) in nearby areas. C^(14) concentration is C_(1) in nearby areas and C_(2) in areas far away. If the age of the fossil is determined to be T_(1) and T_(2) at the places , respectively, then

Carbon -14 used to determine the age of organic material. The procedure is absed on the formation of C^(14) by neutron capture iin the upper atmosphere. ._(7)N^(14)+._(0)n^(1) rarr ._(6)C^(14)+._(1)H^(1) C^(14) is absorbed by living organisms during photosynthesis. The C^(14) content is constant in living organism. Once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of C^(14) in the dead being falls due to the decay, which C^(14) undergoes. ._(6)C^(14)rarr ._(7)N^(14)+beta^(c-) The half - life period of C^(14) is 5770 year. The decay constant (lambda) can be calculated by using the following formuls : lambda=(0.693)/(t_(1//2)) The comparison of the beta^(c-) activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method, however, ceases to be accurate over periods longer than 30000 years. The proportion of C^(14) to C^(12) in living matter is 1:10^(12) . What should be the age of fossil for meaningful determination of its age ?