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Standard X

Physics

Nuclear Fission

Uranium ores ...

Question

Uranium ores contain one raidum $- 226$ atom for every $2.8 \times 10^{6}$ uranium-238 atoms. Calculate the half-life of $_{92} U^{238}$ , given that the half-life of $_{88} R a^{226}$ is $1600$ years $(_{88} R a^{226}$ is a decay product of $_{92} U^{238})$ .

1.75 \times 10^{3} y e a r s

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1600 \times \frac{238}{92} y e a r s

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

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1600 \times 238 y e a r s

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Solution

The correct option is C $4.5 \times 10^{9} y e a r s$
As the ratio of numbers of radium-226 and uranium-238 is maintained at $N_{r} : N_{u} = 1 : 2.8 \times 10^{6}$ , their respective activities(number of disintegration per second) are equal, $A_{r} = A_{u}$
The activity $A \propto N / t_{1 / 2}$
So, $N_{u} / N_{r} = t_{1 / 2}^{u} / t_{1 / 2}^{r}$
$\Rightarrow t_{1 / 2}^{u} = 1600 \times 2.8 \times 10^{6} = 4.5 \times 10^{9} y e a r s$

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Similar questions

Q. Uranium ores contain one radium

- 226

atoms for every

2.8 \times 10^{6}

uranium

- 238

atoms. Calculate the half - life of

_{92}^{238} U

, given that the half - life of

_{88}^{226} R a

is 1600 years. (

_{88}^{226} R a

is a decay product of

_{92}^{238} U

)

Q. Uranium ores contain one radium

- 226

atoms for every

2.8 \times 10^{6}

uranium

- 238

atoms. Calculate the half - life of

_{92}^{238} U

, given that the half - life of

_{88}^{226} R a

is 1600 years. (

_{88}^{226} R a

is a decay product of

_{92}^{238} U

)

Q. (a) On analysis a sample of uranium ore was found to contain 0.277g of

_{82} P b^{206}

and 1.667g of

_{92} U^{238}

.The half life period of

U^{238}

4.51 \times 10^{9}

year. If all the lead were assumed to have come from decay of

_{92} U^{238}

. What is the age of the earth?
(b) An ore of

_{92} U^{238}

is found to contain

_{92} U^{238}

and

_{92} P b^{206}

in the weight ratio of 1:0.1.The half life period of

_{92} U^{238}

4.5 \times 10^{9}

year. Calculate the age of ore?

_{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

)

Q. In uranium mineral, the atomic ratio

N_{U - 238} / N_{P b - 206}

is nearly equal to one. The age (in years) of the mineral is nearly: (half-life period of U-238 is

4.5 \times 10^{9}

years)

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Uranium ores contain one raidum−226 atom for every 2.8×106 uranium-238 atoms. Calculate the half-life of 92U238, given that the half-life of 88Ra226 is 1600 years (88Ra226 is a decay product of 92U238).

The correct option is C 4.5×109yearsAs the ratio of numbers of radium-226 and uranium-238 is maintained at Nr:Nu=1:2.8×106, their respective activities(number of disintegration per second) are equal, Ar=AuThe activity A∝N/t1/2So, Nu/Nr=tu1/2/tr1/2⇒tu1/2=1600×2.8×106=4.5×109 years

Uranium ores contain one raidum $- 226$ atom for every $2.8 \times 10^{6}$ uranium-238 atoms. Calculate the half-life of $_{92} U^{238}$ , given that the half-life of $_{88} R a^{226}$ is $1600$ years $(_{88} R a^{226}$ is a decay product of $_{92} U^{238})$ .