``^238U`decays to ``^206(Pb)`with a half-life of `4.47xx10^9 y.This happens in a number of steps .Can you justify a single half-life for this chain of processes? A sample of rock is found to contain 2.00mg of ``238U`and 0.600 mg of ``206(Pb)`.assuming that all the lead has come form uranium ,find the life of the rock.

U^238 decays to (Pb)^206 with a half-life of 4.47xx10^9 y.The ratio of number of nuclei of pb to u 238 after a time of 1.5 .10 power 9 years given 21/3=1.26

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

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

A sample of U^(238) (half-line = 4.5xx10^(9) yr ) ore is found to contain 23.8g of U^(238) and 20.6g of Pb^(206) Calculate the age of the ore.

A sample of U^(238) (t_(1//2)= 4.5 xx 10^(9) yrs) is found to contains 23.8g U^(238) and 20.6g Pb^(206) , the age of ore is

.^(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)) .

.^(238)U decays with a half-life of 4.5 xx10^(9) years, the decay series eventaully 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 prodduced by the decay of .^(238)U and that all other half-lives on the chain are negligilbe, the age of the rock sample is (1n 1.0058 =5.78 xx10^(-3)) .