Calculate the 147Sm144Nd ratio of a rock containing 183 ppm

Calculate the 147Sm/144Nd ratio of a rock containing 1.83 ppm Sm and 5.51 ppm Nd. The atomic weight of Sm is 150.36 and that of Nd is 144.24. The isotopic percentage abundance of 147Sm is 15.0 and that of 144Nd is 23.954.

Solution

1.83 ppm Sm means 1.83 x10^ -6 gm of Sm per 1 gm rock

moles of Sm = (1.83 x10^ -6/150.36) = 0.01217 x10^ -6,

moles of Nd = ( 5.51x10^ -6/144.24) = 0.0382 x10^ -6 ,

considering % abundance

147 Sm present = (15/100) x 0.01217 x10^ -6 = 0.0018255 x10^ -6 ,

144 Nd present = ( 23.954/100) x ( 0.0382x10^ -6) = 0.00915 x10^ -6,

147 Sm/144Nd = (0.0018255 x10^ -6/0.00915 x10^ -6 )

= 0.1995