00:02
Okay, for the given problem, we are provided with the age of metrival, which is 2 .7, multiplied 10 raise to power 9 years, while the half -life of thalium 232 -thi -h, the upper -in bracket, is 1 .39, 10 -rays -to -power -t, 10 -6 -2 -2 -4 -2 .2.
00:35
10 years.
00:38
Let the mass ratio between lead and thallium be x.
00:42
So it will be 208 lead divided by 232 thalium be x.
00:54
Therefore we know that 208 lead equals x 232 of thalium let this equation be marked as equation a we know that for nucleotide that decay constant can be defined as okay so decay constant can be defined as lambda equals 1 by t time natural log number of thorium atoms initially present so thh atoms in present divided by number of thorium atoms remained after time t.
01:58
Let this equation be marked as equation number one.
02:02
Substituting the values into the equation we have lambda equals 0 .693 divided by t half let this be equation number two from formula from from from formula number two we can get the decay constant lambda so the decay constant of lambda will be equal to 0 .69 divided by 1 .39 multiply 10 raise to power 10 years okay now using our equation number one and substituting the lambda value so lamb lambda equals 1 by t natural log initially present divided by initially number of atom present minus the number of lead atoms formed will give us the value of thorium atom remained so substituting the values, we have natural log.
03:50
We have 1 divided by 2 .7, 10 raise to power 10, natural log, thorium 232 divided by 232 thorium minus 208 lead atoms...