हिंदी

A sample of old wood shows 7.0 dps/g. If the fresh sample of tree shows 16.0 dps/g, how old is the given sample of wood? (Half-life of 14C is 5730 y)

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प्रश्न

A sample of old wood shows 7.0 dps/g. If the fresh sample of tree shows 16.0 dps/g, how old is the given sample of wood? (Half-life of 14C is 5730 y)

संख्यात्मक
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उत्तर

Given: t1/2 of 14C = 5730 y

`(- "dN"_0)/"dt" = 16`dps/g

`(- "dN")/"dt"` = 7 dps/g

To find: Age of sample of wood

Formulae: 

  1. `"t"_(1//2) = 0.693/lambda`
  2. `lambda = 2.303/"t" log_10 ("N"_0/"N")`

Calculation: 

`lambda = 0.693/"t"_(1//2) = 0.693/5730 = 1.209 xx 10^-4 "y"^-1`

`"t" = 2.303/lambda log_10 ("N"_0/"N")`

`= 2.303/(1.209 xx 10^-4) log_10 (16.0/7.0)`

= 6839 y

The age of sample of old wood is 6839 y.

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अध्याय 13: Nuclear Chemistry and Radioactivity - Exercises [पृष्ठ २०३]

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बालभारती Chemistry [English] Standard 11 Maharashtra State Board
अध्याय 13 Nuclear Chemistry and Radioactivity
Exercises | Q 4. (M) | पृष्ठ २०३

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