Uranium-235 (u-235) radioactively decays into lead-207 (pb-207). i'm measuring the number of atoms in a mineral, and i find that there are 500 atoms of u-235 and 1500 atoms of pb-207. if we assume the mineral contained only parent and no daughter, then how many atoms of these elements were in the mineral when the rock first formed?

It contains 192,500atoms of uranium-235 ,zero atom of lead.

Explanation: Radioactivity decay of uranium i.e splitting of uranium into half it's size until it achieve stability while emmiting either alpha particle or beta Ray's eventually ends with stable lead(Pb207),in other words,uranium will have undergo seven(7) alpha decays to achieve a lead 207status ,an alpha particle is represented by an He atom which means a mass number of 4,so 235-(4×7)=207.in the begining ,we are told that only parent elements exists in the mineral,so only uranium will be present,taking it backwards from 1500 atoms of Pb and noting that in each of the decay ,the parent loses half it's atom,then the calculation will be 1500_3000_6000_12,000_24,000_48,000_96,000_192,000+500 initial uranium which gives a total of 192,500atoms present in the mineral at the initial measurement.

i'm not sure

explanation:

i'll try whatever you got

the correct answer is:

a. formation of precipitation