Home
Class 12
CHEMISTRY
An old piece of wood has 25.6% as much C...

An old piece of wood has 25.6% as much `C^(14)` as ordinary wood today has. Find the age of the wood. Half-life period of `C^(14)` is 5760 years?

Text Solution

Verified by Experts

Suppose the amount of `C^(14)` present in the wood originally (i.e., the same which the wood today has) `= a`. Then the amount of `C^(14)` present now in the old wood
`(25.6)/(100) a = 256a`
The time `t` in which `C^(14)` changed from `a` to `0.256a` will then be given by
`t = (2.303)/(K) "log" (a)/(0.256a)`
But `K = (0.693)/(t_(1//1)) = (0.693)/(5760) = 1.203 xx 10^(-4) "year"^(-1)`
`:. t = (2.303)/(1.203 xx 10^(-4)) "log" (1)/(0.256) = 11329` years
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex6.1 Objective|15 Videos
  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex6.2 Objective|8 Videos
  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|13 Videos
  • NCERT BASED EXERCISE

    CENGAGE CHEMISTRY ENGLISH|Exercise Nuclear Chemistry (NCERT Exercise)|29 Videos
  • ORGANIC COMPOUNDS WITH FUNCTIONAL GROUP

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Analytical And Descriptive|24 Videos

Similar Questions

Explore conceptually related problems

The amount of ._(6)C^(14) isotope in a piece of wood is found to be one-fifth of that present in a fresh piece of wood. Calculate the age of wood (Half life of C^(14) = 5577 years)

The ._(6)C^(14) and ._(6)C^(12) ratio in a piece of woods is 1//16 part of atmosphere. Calculate the age of wood. t_(1//2) "of" C^(14) is 5577 years?

A piece of wood from an archaeological source shows a .^(14)C activity which is 60% of the activity found in fresh wood today. Calculate the age of the archaeological sample. ( t_(1//2) for .^(14)C = 5570 year)

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.

The fossil bone has a .^(14)C : .^(12)C ratio, which is [(1)/(16)] of that in a living animal bone. If the half -life of .^(14)C is 5730 years, then the age of the fossil bone is :

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 ?

Equal masses of two samples of charcoal A and B are burnt separately and the resulting carbon dioxide are collected in two vessels. The radioactivity of ^14 C is measured for both the gas samples. The gas from the charcoal A gives 2100 counts per week and the gas from the charcoal A gives 2100 counts per week and the gas from the charcoal B gives 1400 counts per week. Find the age difference between the two samples. Half-life of ^14 C = 5730 y .

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)?

In a sample of wood, the reading of a counter is 32 dpm and in a fresh sample of tree it is 122dpm . Due to error counter gives the reading 2 dpm in absence of .^(14)C . Half life of .^(14)C is 5770 years . The approximate age (in years) of wood sample is :

CENGAGE CHEMISTRY ENGLISH-NUCLEAR CHEMISTRY-Solved Example
  1. In the given radioactive disintegration series, .90^(232)Th to (2)^(...

    Text Solution

    |

  2. The actual atomic mass of .(20)Ca^(40) is 39.96259 amu. Find the bindi...

    Text Solution

    |

  3. 14 g of a radioactive substance decays to 7 g in 20 min. Will the time...

    Text Solution

    |

  4. The final product of U^(238) is Pb^(206). A sample of pitchblende cont...

    Text Solution

    |

  5. An old piece of wood has 25.6% as much C^(14) as ordinary wood today h...

    Text Solution

    |

  6. The half-life of cobalt-60 is 5.26 years. Calculate the percentage act...

    Text Solution

    |

  7. The activity of 1 g radium is found to be 0.5. Calculate the half-life...

    Text Solution

    |

  8. It is found that 3.125 xx 10^(-8) g atoms of Rn exist in equilibrium w...

    Text Solution

    |

  9. What mass of C^(14) with t(1//2) = 5730 years has activity equal to cu...

    Text Solution

    |

  10. The disintegration rate of a certain radioactive sample at any instant...

    Text Solution

    |

  11. A radioisotope .(Z)A^(m) (t(1//2) = 10 days) decays to give .(z - 6)B^...

    Text Solution

    |

  12. At radioactive equilibrium, the ratio between two atoms of radioactive...

    Text Solution

    |

  13. The mean lives of a radioactive substance are 1620 years and 405 years...

    Text Solution

    |

  14. Calculate the effective neutron capture radius of a nucleus having a c...

    Text Solution

    |

  15. A 0.2 mL sample of a solution containing 1.0xx10^(-7) curie of .1H^3 ...

    Text Solution

    |

  16. A sample of .(53)I^(131), as iodide ion, was administered to a patient...

    Text Solution

    |

  17. 1 mg of Th emits 22 alpha-particles per unit solid angle per minute. C...

    Text Solution

    |

  18. A solution contains 1 mCi of L-phenylalanine C^(14) labelled in 2.0 mL...

    Text Solution

    |