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Standard XII

Chemistry

Ideal Gas Equation

In a uranium ...

Question

In a uranium ore the ratio of $U^{238}$ to $P b^{206}$ nuclei is 1:15. Evalute the age of ore, assuming $P b^{206}$ to be the final, initially only uranium was present (given ratio is of number of nuclei).
(Half life= $4.5 \times 10^{9}$ years)

4.5 \times 10^{3}

years

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13.5 \times 10^{3}

years

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9 \times 10^{3}

years

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18 \times 10^{3}

years

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Solution

The correct option is D $18 \times 10^{3}$ years
$N = N_{o} e^{- λ t}$
$U ⟶ P b$
$t = 0$ $N_{0}$ $0$
$t = t$ $N$ $N_{0} - N$
1 $15$

ratio $\frac{N_{o} - N}{N} = 15$
$N_{0} - N = 15 N$
$16 N = N_{o}$
if $N_{o} = 16 N = 1$
This means it has taken 4 half lives.

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Similar questions

Q. In an ore containing uranium, the ratio of

U^{238}

P b^{206}

3

. Calculate the age of the ore, assuming that all the leas present in the ore is the final stable product of

U^{238}

. Take the half-life of

U^{238}

to be

4.5 \times 10^{9}

yr.

Q. In an ore containing uranium the ratio of

U^{238}

P b^{206}

nuclei in

3

. Calculate the age of ore, assuming that all the lead present in the ore is the final stable product of

U^{238} . t_{1 / 2}

for

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

_{921} U^{238}

by successive radioactivity decays changes to

_{82} P b^{206}

.A sample of uranium ore was analyzed and found to contain 1.0g of

U^{238}

and

0.1

g of

P b^{206}

. Assuming that all the

P b^{206}

had accumulated due to decay of

U^{238}

,find out the age of the ore ?(Half life of

U^{238} = 4.5 * 10^{9} y e a r s) .

Q. A sample of

U^{238}

(half-life

= 4.5 \times 10^{9}

years) ore is found to contain

23.8 g

U^{238}

and

20.6 g

P b^{206}

. The age of the ore is:

_{92} U^{238}

by successive radioactive decays changes to

_{82} {Pb}^{206}

. A sample of uranium ore was analyzed and found to contain 1.0 g of

U^{238}

and 0.1 g of

P b^{206}

. Assuming that all the

P b^{206}

had accumulated due to decay of

U^{238}

, find out the age of the ore. (Half life of

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

)

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In a uranium ore the ratio of U238 to Pb206 nuclei is 1:15. Evalute the age of ore, assuming Pb206 to be the final, initially only uranium was present (given ratio is of number of nuclei). (Half life= 4.5×109 years)

The correct option is D 18×103 yearsN=Noe−λt U⟶ Pb t=0 N0 0 t=t N N0−N 1 15 ratio No−NN=15 N0−N=15N 16N=No if No=16N=1 This means it has taken 4 half lives.

In a uranium ore the ratio of $U^{238}$ to $P b^{206}$ nuclei is 1:15. Evalute the age of ore, assuming $P b^{206}$ to be the final, initially only uranium was present (given ratio is of number of nuclei).
(Half life= $4.5 \times 10^{9}$ years)

The correct option is D $18 \times 10^{3}$ years
$N = N_{o} e^{- λ t}$
$U ⟶ P b$
$t = 0$ $N_{0}$ $0$
$t = t$ $N$ $N_{0} - N$
1 $15$

ratio $\frac{N_{o} - N}{N} = 15$
$N_{0} - N = 15 N$
$16 N = N_{o}$
if $N_{o} = 16 N = 1$
This means it has taken 4 half lives.