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Byju's Answer

Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

A sample of w...

Question

A sample of wooden aircraft is found to undergo 9dpm $g^{- 1}$ of $^{14} C$ . The half-life of $_{6}^{14} C$ is 5730 year and rate of disintegration of wood recently out down is 15 dpm per g of $_{6}^{14} C$ . The approximate age of aircraft is:

4224.35 year

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5234.31 year

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4224.47 year

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Solution

The correct option is C 4224.47 year
$t = \frac{2.303 \times t_{1 / 2}}{0.693} \times l o g \frac{N_{0}}{N} = \frac{2.303 \times 5730}{0.693} \times l o g \frac{15}{9} = 4224.47 y e a r$
Thus, the approximate age of aircraft is 4224.47 years.

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A sample of wooden aircraft is found to undergo 9dpm g−1 of 14C. The half-life of 146C is 5730 year and rate of disintegration of wood recently out down is 15 dpm per g of 146C. The approximate age of aircraft is:

The correct option is C 4224.47 year t=2.303×t1/20.693×logN0N=2.303×57300.693×log159=4224.47yearThus, the approximate age of aircraft is 4224.47 years.

A sample of wooden aircraft is found to undergo 9dpm $g^{- 1}$ of $^{14} C$ . The half-life of $_{6}^{14} C$ is 5730 year and rate of disintegration of wood recently out down is 15 dpm per g of $_{6}^{14} C$ . The approximate age of aircraft is:

The correct option is C 4224.47 year
$t = \frac{2.303 \times t_{1 / 2}}{0.693} \times l o g \frac{N_{0}}{N} = \frac{2.303 \times 5730}{0.693} \times l o g \frac{15}{9} = 4224.47 y e a r$
Thus, the approximate age of aircraft is 4224.47 years.