A piece of wood from the ruins of an ancient building was found to have a 14 C activity of 12 disintegrations 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 half-life of 14 C is 5760 yr.

This is a Short Answer Type Questions as classified in NCERT ExemplarExplanation- R= 12dis/min per gm and R0= 16dis/min per gm T1/2= 5760yrR= e - λ t e λ t By taking log λ t l o g e = l o g R 0 R λ t = log1016/12 × 2.303 t = 2.303 ( l o g 4 - l o g 3 ) λ = 2.303 (0.6020-4.771)5760/0.6931 = 2391.20yr

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Nuclei Physics Ncert Solutions Class 12th Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

A piece of wood from the ruins of an ancient building was found to have a ¹⁴C activity of 12 disintegrations per minute per gram of its carbon content. The ¹⁴C 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 half-life of ¹⁴C is 5760 yr.

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alok kumar singh | Contributor-Level 10

4 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- R= 12dis/min per gm and R₀= 16dis/min per gm T_1/2= 5760yr

R= $e^{- λ t}$ $e^{λ t}$

By taking log

$λ t l o g e = l o g \frac{R_{0}}{R}$

$λ t$ = log₁₀16/12 $\times 2.303$

$t = \frac{2.303 (l o g 4 - l o g 3)}{λ}$

= 2.303 (0.6020-4.771)5760/0.6931

= 2391.20yr

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