1 g of $^{14} C$ from an archaeological sample gives 9.4 counts $m i n^{- 1}$ . 1 g of $^{14} C$ from a recently cut tree given 15.3 counts $m i n^{- 1}$ under the same conditions. How old is the archaeological sample? ( $T_{50}$ of $^{14} C$ is 5730 years).

4028.62 years

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7628.62 years

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4528.62 years

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None of these

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Solution

The correct option is A 4028.62 years
We know, $λ = \frac{0.693}{t_{1 / 2}}$ where, $t_{1 / 2}$ = half life-time.
So, for first order reaction, the equation is :

$λ = \frac{2.303}{t} l o g \frac{C_{0}}{C}$

Substituting the value for $λ$ , we get

$\frac{0.693}{5730} = \frac{2.303}{t} l o g \frac{15.3}{9.4}$

So, $t = 4028.62 y e a r s$

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Similar questions

Q. An old sample of wood from an archaeological excavation was found to have a radioactivity of about 140.8 counts per minute due to

^{14} C (t_{1 / 2} = 5730

years) as against 281.6 counts per minute for that of a freshly cut piece of wood. What is the age of the old sample of wood (in millennia, 1 millennium= 1000 years)?

Q. A piece of wood from an archaeological sample has

5.0 c o u n t s {m i n}^{- 1}

per gram of C-14, while a fresh sample of wood has a count of

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. If half-life of C-14 is

5770

yrs, the age of the archaeological sample is:

The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.

Q. When charcoal is prepared from a living tree, it shows a disintegration rate of

15.3

disintegrations of

^{14} C

per gram per minute. A sample from an ancient piece of charcoal shows

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activity to be

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5730 years and ln 0.8039 = - 0.2182)

The half-life for radioactive decay of $^{14} C$ is 5730 yr. An archaeological artifact containing wood had only $80 %$ of the $^{14} C$ found in a living tree. Estimate the age of the sample.

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1 g of 14C from an archaeological sample gives 9.4 counts min−1. 1 g of 14C from a recently cut tree given 15.3 counts min−1 under the same conditions. How old is the archaeological sample? (T50 of 14C is 5730 years).

The correct option is A 4028.62 yearsWe know, λ=0.693t1/2 where, t1/2 = half life-time.So, for first order reaction, the equation is :λ=2.303tlogC0CSubstituting the value for λ, we get0.6935730=2.303tlog15.39.4So, t=4028.62years

1 g of $^{14} C$ from an archaeological sample gives 9.4 counts $m i n^{- 1}$ . 1 g of $^{14} C$ from a recently cut tree given 15.3 counts $m i n^{- 1}$ under the same conditions. How old is the archaeological sample? ( $T_{50}$ of $^{14} C$ is 5730 years).