The age of a sample of wood in years- if only 116 part of original 14C is present in its piece- is- Half life of 14C is 5580 years

MM_0=116=12^n 2^n=16=2^4 n=4 The age of the sample of wood is, nbsp;t=nT=4 × 5580 t=22320 years Hence, (D) is the correct answer. ...

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Standard IX

Chemistry

Radiocarbon Dating

The age of a ...

Question

The age of a sample of wood (in years), if only $\frac{1}{16}$ part of original $^{14} C$ is present in its piece, is, (Half life of $^{14} C$ is $5580$ years)

5580

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11160

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16740

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22320

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Solution

The correct option is D $22320$
$\frac{M}{M_{0}} = \frac{1}{16} = \frac{1}{2^{n}}$

$2^{n} = 16 = 2^{4}$
$n = 4$

The age of the sample of wood is,
$t = n T = 4 \times 5580$
$t = 22320 years$

Hence, $(D)$ is the correct answer.

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The age of a sample of wood (in years), if only 116 part of original 14C is present in its piece, is, (Half life of 14C is 5580 years)

The correct option is D 22320MM0=116=12n 2n=16=24 n=4 The age of the sample of wood is, t=nT=4×5580 t=22320 years Hence, (D) is the correct answer.

The age of a sample of wood (in years), if only $\frac{1}{16}$ part of original $^{14} C$ is present in its piece, is, (Half life of $^{14} C$ is $5580$ years)

The correct option is D $22320$
$\frac{M}{M_{0}} = \frac{1}{16} = \frac{1}{2^{n}}$

$2^{n} = 16 = 2^{4}$
$n = 4$

The age of the sample of wood is,
$t = n T = 4 \times 5580$
$t = 22320 years$

Hence, $(D)$ is the correct answer.