Plot the following data for mineral separates from a meteorite (LEW 86010) in a
ID: 108108 • Letter: P
Question
Plot the following data for mineral separates from a meteorite (LEW 86010) in a ^142 Nd/^144 Nd (or epsilon_142Nd) vs. ^144Sm/^l44 Nd diagram, including error bars. Explain the significance of this fossil isochron diagram. Determine the initial ^146 Sm/^144 Sm at the time of formation of this meteorite. Explain how this ratio may be used as a chronometer for early solar system processes. Assuming the solar system initial ^146 Sm/^144 Sm is 0.008 at time zero, calculate the time elapsed since the beginning of the solar system until this sample has formed. The half-life of ^146Sm is 103Ma. Use the following constants to solve these problems: A =0.581 times 10^-10 y^-1. Lambda_p = 4.962 times 10^-10 y^-1, lambda = 5.543 times 10^-10 y^-1, Ab^40 K = 0.01167 % (atom), At. Wt. K = 39.0983.Explanation / Answer
When142Nd/144Nd vs 144Sm/144Nd, when 142Nd/144Nd increases linearly,the value 144Sm/144Nd shows a decrease and then an increase which gives an appearence like a parabola. The value 142Nd/144Nd is taken along x axis while 144Sm/144Nd is taken along y axis and the resultant graph is in the form of parabola and the error bars are plotted as small points on the parabolic structure.
Initially the value of 146Sm/144Sm will be small and as the age of fossil fuels rise,the value of ratio will also rise.This ratio is useful is spotting the age of planets in the solar system. That is,using the date and age of formation of fossil fuels,the age of earth can be found because the fossil fuels buried inside the ground will be millions if years old and using earth's age other plantes age,their surface area can be determined approximately.
Initial 146Sm/144Sm value=0.008
The half life of 146Sm=103Ma
Lamda e=0.581*10^-10y-1
lamda p=4.962*10^-10y-1
lamda=5.543*10^-10y-1
Time elapse=(half life*initial value)/(lamda)
=(103*0.008)/(5.543*10^-10)
=14.86*10^8yr