A piece of bone from an archaeological site is found to give a count rate of $15$ counts per minute. A similar sample of fresh bone gives a count rate of $19$ counts per minute. Calculate the age of the specimen. Given $T_{1 / 2} = 5570 y e a r s$

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Answer 1

Count rate of fresh sample $N_{0} = 19$ counts/minute
Count rate of bone $N = 15$ counts/minute
$T_{1 / 2} = 5570$ years, Age of the sample, $t = ?$
$N = N_{0} e^{- λ} t$
$λ = \frac{0 . 6 9 3 1}{5 5 7 0}$
$15 = 19 e^{- λ t}$ (or) $e^{λ t} = \frac{1 9}{1 5}$
$lo g e \frac{1 9}{1 5} = λ t$ (or) $t = lo g e \frac{1 9}{1 5} \times \frac{1}{λ}$
$t = \frac{5 5 7 0}{0 . 6 9 3 1} \times 2.3026 \times lo g_{10} \frac{1 9}{1 5}$
$= \frac{5 5 7 0}{0 . 6 9 3 1} \times 2.3026 \times 0.1026$
$t = 1899 y e a r s$ .

Answer 2

Count rate of fresh sample

N_{0} = 19

counts/minute

Count rate of bone

N = 15

counts/minute

T_{1 / 2} = 5570

years, Age of the sample,

t = ?

N = N_{0} e^{- λ} t

λ = \frac{0 . 6 9 3 1}{5 5 7 0}

15 = 19 e^{- λ t}

(or)

e^{λ t} = \frac{1 9}{1 5}

lo g e \frac{1 9}{1 5} = λ t

(or)

t = lo g e \frac{1 9}{1 5} \times \frac{1}{λ}

t = \frac{5 5 7 0}{0 . 6 9 3 1} \times 2.3026 \times lo g_{10} \frac{1 9}{1 5}

= \frac{5 5 7 0}{0 . 6 9 3 1} \times 2.3026 \times 0.1026

t = 1899 y e a r s

.