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A 25.0 g piece of charcoal is found in some ruins of an ancien

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A $25.0$ g piece of charcoal is found in some ruins of an ancient city. The sample shows a $^{14} C$ activity of 250 decays / min. How long has the tree this charcoal came from been dead? Given : the half-life of $^{14} C$ is $5730 yr$ . The ratio of $^{14} C$ and $^{12} C$ in the living sample is $1.3 \times 1 0^{- 12}$ .

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25.0

g piece of charcoal is found in some ruins of an ancient city. The sample shows a

^{14} C

activity of 250 decays / min. How long has the tree this charcoal came from been dead? Given : the half-life of

^{14} C

5730 yr

. The ratio of

^{14} C

and

^{12} C

in the living sample is

1.3 \times 1 0^{- 12}

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

The decay constant

λ

^{14} C

will be,

λ = \frac{0.693}{t _{1/2}} = \frac{0.693}{( 5730 ) ( 365 \times 24 \times 60 \times 60 )} = 3.84 \times 1 0^{- 12} s^{- 1}

The number of

^{14} C

nuclei can be calculated in two steps.
First, the number of

^{12} C

nuclei in

25.0 g

of carbon is,

N (^{12} C) = (6.02 \times 1 0^{23}) (\frac{25}{12}) = 1.25 \times 1 0^{24}

The ratio of

^{14} C

^{12} C

in the live sample was

1.3 \times 1 0^{- 12}

, we see that number of

^{14} C

nuclei in

25.0 g

before decay was,

N_{0} (^{14} C) = (1.3 \times 1 0^{- 12}) (1.25 \times 1 0^{24}) = 1.6 \times 1 0^{12}

Hence, the initial activity of the sample was,

R_{0} = λ N_{0} = (3.84 \times 1 0^{- 12}) (1.6 \times 1 0^{12}) = 6.14 decay / s = 370 decay / min

Activity at the time of observation is given as, 250 decay/min.
From the equation,

R = R_{0} e^{- λ t}

we have,

λ t = ln (\frac{R _{0}}{R}) t = \frac{1}{λ} ln (\frac{R _{0}}{R})

or Substituting the values, we have

t = \frac{1}{3.84 \times 1 0 ^{- 12}} ln (\frac{370}{250}) = 1.0 \times 1 0^{11} s = 3200 yr

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Question Text	A $25.0$ g piece of charcoal is found in some ruins of an ancient city. The sample shows a $^{14} C$ activity of 250 decays / min. How long has the tree this charcoal came from been dead? Given : the half-life of $^{14} C$ is $5730 yr$ . The ratio of $^{14} C$ and $^{12} C$ in the living sample is $1.3 \times 1 0^{- 12}$ .
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	90

Text solutionVerified

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A 25.0 g piece of charcoal is found in some ruins of an ancient city. The sample shows a 14C activity of 250 decays / min. How long has the tree this charcoal came from been dead? Given : the half-life of 14C is 5730yr. The ratio of 14C and 12C in the living sample is 1.3×10−12.

Text solutionVerified

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