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Radiocarbon Dating

A sample of ...

Question

A sample of $U^{238} (t_{1 / 2} = 4.5 \times 10^{9} y e a r s)$ ore is found to contain 23.8 g of $U^{238}$ and 20.6 g of $P b^{206}$ . Calculate rate constant of the rock.

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Solution

Its a 1st Order Reaction

$k = \frac{0.693}{t_{1 / 2}}$

$k = 1.54 \times 10^{- 10} {y r s}^{- 1}$

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Similar questions

U^{238}

= 4.5 \times 10^{9}

23.8 g

U^{238}

20.6 g

P b^{206}

X \times 10_{9}

2 X

U^{238}

4.5 \times 10^{9}

U^{238}

P b^{206} .

U^{238} ((h a l f - l i f e = 4.5 \times 10^{9} y r)

U^{238}

P b^{206}

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4.47 \times 10^{9}

U^{238}

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_{92} U^{238}

_{82} {Pb}^{206}

U^{238}

P b^{206}

P b^{206}

U^{238}

U^{238} = 4.5 \times 10^{9} y e a r s

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