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The final product of U^(238) is Pb^(206)...

The final product of `U^(238)` is `Pb^(206)`. A sample of pitchblende contains `0.0453 g` of `Pb^(206)` for every gram of `U^(238)` present in it. Supposing that the mineral pitchblende formed at the time of formation of the earth did not contain any `Pb^(206)`, calculate the age of the earth (half-life period of `U^(238) = 4.5 xx 10^(9)` years).

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To calculate the age of the Earth based on the decay of uranium-238 to lead-206, we can follow these steps: ### Step 1: Understand the relationship between uranium-238 and lead-206 The decay of uranium-238 (U-238) leads to the formation of lead-206 (Pb-206). In the given problem, we know that for every gram of U-238, there is 0.0453 grams of Pb-206 present. ### Step 2: Determine the initial amount of uranium-238 Let’s assume we have 1 gram of U-238 currently. Since Pb-206 was not present when the Earth formed, the initial amount of U-238 would be the current amount plus the amount that has decayed into Pb-206. ...
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In a smaple of rock, the ration .^(206)Pb to .^(238)U nulei is found to be 0.5. The age of the rock is (given half-life of U^(238) is 4.5 xx 10^(9) years).

A sample of uranium mineral was found to contain Pb^(208) and U^(238) in the ratio of 0.008 : 1. Estimate the age of the mineral (half life of U^(238) is 4.51 xx 10^(9) years).

(a) On analysis a sample of uranium ore was found to contain 0.277g of ._(82)Pb^(206) and 1.667g of ._(92)U^(238) . The half life period of U^(238) is 4.51xx10^(9) year. If all the lead were assumed to have come from decay of ._(92)U^(238) , What is the age of earth? (b) An ore of ._(92)U^(238) is found to contain ._(92)U^(238) and ._(82)Pb^(206) in the weight ratio of 1: 0.1 The half-life period of ._(92)U^(238) is 4.5xx10^(9) year. Calculate the age of ore.

In an ore containing Uranium, the ratio of U^(238) to Pb^(206 nuceli is 3 . Calculate the age of the ore, assuming that alll the lead present in the ore is the final stable, product of U^(238) . Take the half-like of U^(238) to be 4.5 xx 10^(9) years. In (4//3) = 0.288 .

A rock is 1.5 xx 10^(9) years old. The rock contains .^(238)U which disintegretes to form .^(236)U . Assume that there was no .^(206)Pb in the rock initially and it is the only stable product fromed by the decay. Calculate the ratio of number of nuclei of .^(238)U to that of .^(206)Pb in the rock. Half-life of .^(238)U is 4.5 xx 10^(9). years. (2^(1/3) = 1.259) .

In an ore containing uranium, the ratio of ^238U to 206Pb nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of ^238U . Take the half-life of ^238U to be 4.5xx10^9 years.

A sample of U^(238) ("half life" = 4.5 xx 10^(9)yr) ore is found to contain 23.8 g" of " U^(238) and 20.6g of Pb^(206) . Calculate the age of the ore

.^(238)U decays with a half-life of 4.5 xx10^(9) years, the decay series eventually ending at .^(206)Pb , which is stable. A rock sample analysis shows that the ratio of the number of atoms of .^(206)Pb to .^(238)U is 0.0058. Assuming that all the .^(206)Pb is produced by the decay of .^(238)U and that all other half-lives on the chain are negligible, the age of the rock sample is (ln 1.0058 =5.78 xx10^(-3)) .

The isotope of U^(238) and U^(235) occur in nature in the ratio 140:1 . Assuming that at the time of earth's formation, they were present in equal ratio, make an estimate of the age of earth. The half lives of U^(238) and U^(235) are 4.5xx10^9 years and 7.13xx10^8 years respectively. Given log_(10)140=2.1461 and log_(10)2=0.3010.

In a sample of rock, the ratio of number of ""^(206)Pb to ""^(238)U nuclei is found to be 0.5. The age of the rock is (18//n)xx10^(9)(ln((3)/(2)))/(ln2) year (Assume that all the Pb nuclides in he rock was produced due to the decay of Uranium nuclides and T_(1//2)(""^(238)U)=4.5 xx 10^(9) year). Find n.

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