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Standard XII

Physics

Alpha Decay

A piece of bo...

Question

A piece of bone of an animal from a ruin is found to have $^{14} C$ activity of $12$ disintegrations per minute per gm of its carbon content. The $^{14} C$ activity of a living animal is $16$ disintegrations per minute per gm. Nearly, how long ago did the animal die? (Given half life of $^{14} C$ is $t_{\frac{1}{2}} = 5760 y e a r s)$

1672

years

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2391

years

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3291

years

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4453

years

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Solution

The correct option is A $2391$ years
We know that:
$A = A_{o} e^{- λ t}$
Therefore, $12 = 16 e^{- λ t}$
$t = \frac{l o g (4 / 3)}{λ}$
Also, $λ = \frac{l o g 2}{t_{(1 / 2)}}$
Solving above equations, we get:
$t = 2391 y e a r s$

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A piece of bone of an animal from a ruin is found to have 14C activity of 12 disintegrations per minute per gm of its carbon content. The 14C activity of a living animal is 16 disintegrations per minute per gm. Nearly, how long ago did the animal die? (Given half life of 14C is t12=5760years)

The correct option is A 2391 yearsWe know that:A=Aoe−λtTherefore, 12=16e−λtt=log(4/3)λAlso, λ=log2t(1/2)Solving above equations, we get:t=2391 years

The correct option is A $2391$ years
We know that:
$A = A_{o} e^{- λ t}$
Therefore, $12 = 16 e^{- λ t}$
$t = \frac{l o g (4 / 3)}{λ}$
Also, $λ = \frac{l o g 2}{t_{(1 / 2)}}$
Solving above equations, we get:
$t = 2391 y e a r s$