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

7997.2

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57, 570

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11, 540

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15140

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Solution

The correct option is C $11, 540$
It is given in the question that, the counter gives the reading 2 dpm in absence of $^{14} C$
So,
The actual correct reading for the counter in sample of wood= $(32 - 2) = 30 dpm$
The actual correct reading for the fresh sample of tree= $(122 - 2) = 120 dpm$
To calculate age,
$λ \times t = 2.303 \times l o g \frac{N_{0}}{N_{t}}$
$\frac{0.693}{5770} \times t = 2.303 \times l o g \frac{120}{30}$

$t =$ $11540 y e a r s$ (approximately)

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

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