Geol. 656 Isotope Geochemistry

Geol. 656 Isotope GeochemistryProblem Set 4 Solutions Due March 6, 20001. Assuming that the decay constant of 36 Cl is 2.3 × 10 -6 yr -1 , and that the cosmogenic productionrate of 36 Cl in a rock is 51 atoms/yr/gram, calculate the exposure age of the two boulders below:N (atoms 36 Cl/g)Boulder 1 5.19 × 10 6Boulder 2 1.28 × 10 6From equation 13.11, we can easily derived the following:⎛ λ ⎞−ln 1 −N ⎝ P ⎠t =λFrom this equation, we calculate the age of boulders 1 and 2 to be 116,000 and 25.900 yearsrespectively.2. Show that the equation:87 86Sr/ Sr true = 87’’86’ 88’’ 87’ × 8.37521corrects for both mass fractionation and differences in amplifier gain between two faraday cups.87" is the measured signal intensity of 87 Sr in cup 2, 87' is the measured signal intensity in cup 1.Let's define gain factors for each of the two faraday cups as G' and G"". We will also definethe fractionation factor between two isotopes a and b as ƒa/b, i.e., a/b) true =a/b) measured ׃a/bThen we can express the 87" and 87' ratios as:86' 88"87/86 true= G"G’ f 7/6 and 87/88 true = G’G" f 7/88.372521 is 1/0.11940, the 88/86) true ratio.Substituting into the equation above, we have87 86Sr/ Srtrue = 87 G"86 trueG’ f 7/6which reduces to:87 86Sr/ Srtrue = 87286 truef 7/6f 7/887 G’88 trueG" f 7/8We need now only to show that ƒ 7/6 = 1/ƒ 7/8For power law fractionation:M [1+α] ∆m i,jR C i,j=R i,j8886 trueso thatf =1+α ∆m i,jandf 7/6=1+α ∆m 87,86and f 7/8=1+α ∆m 87,88Since ∆m 87,88 = -∆m 87,86 , thenƒ 7/8 = (ƒ 7/6 ) -1Q.E.D.1

Geol. 656 Isotope GeochemistryProblem Set 4 Solutions Due March 6, 20003. How long will a 40 K + ion spend traversing a 2 m mass spectrometer flight tube if it has beenaccelerated by a potential of 10 kV?We start with:E = eV = 1 2 mv2We can easily solve this equation for v, and from that for the time.E = 10 4 electron volts = 10 4 × 1.60210 × 10 -19 kg m 2 /sec 2m = 40 daltons = 40 × 1.6604345 × 10 -27 kg = 6.642 × 10 -26 kgSolving for v, we find v = 2.196 × 10 5 m/sFor a 2 meter flight path, the flight time is 9.106 µsec.4. What magnetic field (in gauss) is required to steer a 143 Nd atom through a 90° sector, 30 cmradius mass spectrometer when the accelerating potential is set to 10 kV?Equation 15.6a is:mq= 4.825 × 10-5Br2 2V15.6aWith B is in gauss, r in cm, V in volts, m in daltons, and e in units of electronic charge.Solving for B, we have:B 143.95 mV=r eUsing m = 143 daltons, V = 10000 V; e = 1, r = 30, we find that:B = 5738.5 gauss5. If the magnetic field in the mass spectrometer described above is set to intensity you havecalculated, what would be the spacing in mm between a faraday cup collecting 143 Nd + and onecollecting 144 Nd + ?R 2R 1Rearranging the equation above, we have:2

Geol. 656 Isotope GeochemistryProblem Set 4 Solutions Due March 6, 2000r = 143.95Bm V eWe solve this to find r 2 = 30.1047, so the collector spacing is 1.047 mm6. Suppose you measure a 87 Sr/ 86 Sr ratio of 0.70734 with a mass spectrometer. To monitor massfractionation, you also measured 86 Sr/ 88 Sr and found a value of 0.11992. Assume the true value of86 Sr/ 88 Sr is 0.11940.a. What is the true 87 Sr/ 86 Sr ratio according to the linear mass fractionation law?87⎛ Sr⎞⎜ 0 7073487 ⎟ = .⎝ ⎠Srmeas⎛According to the linear mass fractionation law: ⎜⎝So:87⎛ Sr⎞⎜ 0 70888087 ⎟ = .⎝ ⎠Srtrue8787Sr⎞Sr87Sr⎟ = ⎛ ⎞187⎠ ⎝ ⎜ ⎜Sr⎠truemeas⎜⎛⎟ +⎝0.11992− 1⎞0.1194 ⎟2⎟⎠⎝ b. What ⎠ ⎝ ⎠trueis the true 87 measSr/ ⎜86 Sr ratio according ⎟ to the power mass fractionation law?⎝⎠For power law fractionation: α = ⎡ ⎣ ⎢RRNuvMuv1 / ∆m uv− 1 / 2⎤⎥ − = ⎡ 0.119401⎦ ⎣ ⎢ ⎤0.11992⎦⎥α = R N 1/∆m uv–1=uv0.11940 –1/2– 1 = 0.00426227R M 0.12042And since R C i,j=R M i,j[1+α] ∆m i,j87⎛ Sr⎞⎜ 0 70887987 ⎟ = .⎝ ⎠Srtrue− 1 = 0.002175193