Question

A rock sample contains traces of 23sU, 235u, 237Th, 208pb, 207Pb, and 205Pb. Careful analysis shows that the ratio of 233U to 2Gs 1.254. (a) Use this information to determine the age of the rock. Age 3,78001852 1years (b) What should be the ratios of 235U to 207Pb and 232Th to 208Pb so that they would yield the same age for the rock7 Neglect the minute amount of the intermediate decay products in the decay chains. Note that this form of multiple dating gives reliable geologic dates. NU235/Npb207-1.001527853X NTh232/Npb208

Answer 1

IF YOU HAVE ANY DOUBTS COMMENT BELOW

ANSWER:

EXPLANATION:

a) t = (1/k) ln( a/a-x)

k for U238 = 0.693/4.47 x10^ 9 = 1.55 x10^ -10 , let a = 238U , x is converted amount of U238 i.e 206 Pb , given a-x/x = 1.164 , x = a/2.164 = 0.4417a , the a-x = (1-0.4417)a = 0.5583a ,

now t = ( 1/1.55 x10^ -10) ln( a/0.5583a)

=3.76 x10^ 9 years

b) k for U235 = 0.693/4.04 x10^ 8 = 1.715 x10^ -9 ,

now 3.76 x10^ 9 = (1/1.715 x10^ -9) ln( a/a-x)

a/a-x = 631.7 = a = 631.7 (a-x) , x = 0.99842a ,

a/x = U235/Pb 207 = 1/(0.99842) = 1.00158,

for 232 Th , k = 0.693/1.41 x10^ 10 = 4.915 x10^ -11

t = 3.76 x10^ 9 = ( 1/4.915x10^ -11) ln( a/a-x)

a/a-x = 1.203 , x = 0.168745,

a/x = Th232/Pb208 = 1/0.168745 = 5.926