SATHEE: Atoms And Nuclei Question 406

Atoms And Nuclei Question 406

Question: In an ore containing uranium, the ratio of $U^{238}$ to $P b^{206}$ nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore in the final stable product of $U^{238}$ . Take the half-life of $U^{238}$ to be $4.5 \times 10^{9}$ ear.

Options:

A) $1.8 \times 10^{9}$ year

B) $2.3 \times 10^{10}$ year

C) $5.1 \times 10^{7}$ year

D) $6.2 \times 10^{6}$ year

Show Answer

Answer:

Correct Answer: A

Solution:

Suppose x is the number of $P b^{206}$ nulei.

The number of $U^{238}$ nuclei will be 3x, Thus $3 x + x = N_{0}$ We know that

$N = N_{0} e^{- λ t}$ or $3 x = 4 x e^{- λ t}$
$∴ e^{λ t} = \frac{4}{3}$

or $t = \frac{I n, 4 / 3}{λ} = \frac{I m 4 / 3}{(0.693 / t_{1 / 2})}$ $= \frac{I n 4 / 3}{(0.693 / 4.5 \times 10^{9})} = 1.868 \times 10^{9}$ years.

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Atoms And Nuclei Question 406

Question: In an ore containing uranium, the ratio of U238 to Pb206 nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore in the final stable product of U238 . Take the half-life of U238 to be 4.5×109 ear.

Options:

Answer:

Solution:

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Question: In an ore containing uranium, the ratio of $U^{238}$ to $P b^{206}$ nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore in the final stable product of $U^{238}$ . Take the half-life of $U^{238}$ to be $4.5 \times 10^{9}$ ear.