In a sample of rock- the ratio of - -206Pb- to - -238U- nuclei is found to be 0-5- The age of the rock is -given half-life of - U-238- is - 4-5- 10-9- years- 2-25- 10-9 year- 4-5- 10-9- ln 3 year- 4-5- 10-9ln -3-2 -ln 2 year- 2-25- 10-9 ln -3-2 - year-

The correct option is C 4.5× 10^9ln (3/2 )/ln 2 year First order decay: #160; N(t) =N_o e^ λ t λ = ln2t_1/2 N(t)/N_o =(1+0.5)/0.5 = 3/2 ...

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Byju's Answer

Standard VII

Chemistry

Sub-Atomic Particles

In a sample o...

Question

In a sample of rock, the ratio of $^{206} P b$ to $^{238} U$ nuclei is found to be 0.5. The age of the rock is (given half-life of $U^{238}$ is $4.5 \times 10^{9}$ years)

2.25 \times 10^{9} y e a r

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4.5 \times 10^{9}

ln 3 year

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4.5 \times 10^{9} \frac{l n (\frac{3}{2})}{l n 2} y e a r

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2.25 \times 10^{9} l n (\frac{3}{2}) y e a r

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Solution

The correct option is C $4.5 \times 10^{9} \frac{l n (\frac{3}{2})}{l n 2} y e a r$
First order decay: $N (t) = N_{o} e^{- λ t}$
$λ = \frac{l n 2}{t_{1 / 2}}$
$N (t) / N_{o} = (1 + 0.5) / 0.5 = 3 / 2$
$1.5 = e^{- λ t}$
$λ t = l n 1.5$
$t = 4.5 \times 10^{9} \frac{l n (3 / 2)}{l n 2}$

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In a sample of rock, the ratio of 206Pb to 238U nuclei is found to be 0.5. The age of the rock is (given half-life of U238 is 4.5×109 years)

The correct option is C 4.5×109ln(32)ln2yearFirst order decay: N(t)=Noe−λtλ=ln2t1/2N(t)/No=(1+0.5)/0.5=3/21.5=e−λtλt=ln1.5t=4.5×109ln(3/2)ln2

In a sample of rock, the ratio of $^{206} P b$ to $^{238} U$ nuclei is found to be 0.5. The age of the rock is (given half-life of $U^{238}$ is $4.5 \times 10^{9}$ years)

The correct option is C $4.5 \times 10^{9} \frac{l n (\frac{3}{2})}{l n 2} y e a r$
First order decay: $N (t) = N_{o} e^{- λ t}$
$λ = \frac{l n 2}{t_{1 / 2}}$
$N (t) / N_{o} = (1 + 0.5) / 0.5 = 3 / 2$
$1.5 = e^{- λ t}$
$λ t = l n 1.5$
$t = 4.5 \times 10^{9} \frac{l n (3 / 2)}{l n 2}$