Question

The \(\frac{^{143}Nd}{^{144}Nd}\ \text{and}\ \frac{^{147}Sm}{^{144}Nd}\) ratios of a rock are 0.516 and 0.389, respectively. The rock evolved as a closed system. As per the exact parent-daughter relationship \(\frac{^{143}Nd}{^{144}Nd}\) equation, the ratio of the rock 4.6 × 10⁹ years ago was ______. (Round off to three decimal places)
(Use decay constant for ¹⁴⁷Sm = 6.54 × 10^-12 y^-1)

Answer 1

Correct Answer: 0.502

Given:
Present ratios for the rock: \(\displaystyle \frac{{}^{143}\mathrm{Nd}}{{}^{144}\mathrm{Nd}}\Big|_{\text{now}}=0.516,\qquad \frac{{}^{147}\mathrm{Sm}}{{}^{144}\mathrm{Nd}}\Big|_{\text{now}}=0.389.\)
Decay constant: \(\lambda_{^{147}\mathrm{Sm}}=6.54\times10^{-12}\,\mathrm{y}^{-1}\). Time interval: \(t=4.6\times10^{9}\,\mathrm{y}\). (Assume a closed system.)  
Step 1 — Parent → daughter mass-balance
If \(N_{P}(0)\) and \(N_{D}(0)\) are the parent (Sm) and daughter (Nd) amounts initially, then after time \(t\): \[ N_P(t)=N_P(0)\,e^{-\lambda t}, \qquad N_D(t)=N_D(0)+\big[N_P(0)-N_P(t)\big]. \] The term in brackets is the amount of parent decayed into daughter. Divide everything by the stable isotope \({}^{144}\mathrm{Nd}\) (which does not change). Using present (now) values and rearranging gives the standard relation \[ \boxed{\; \left(\frac{^{143}\mathrm{Nd}}{^{144}\mathrm{Nd}}\right)_{\!{\rm now}} = \left(\frac{^{143}\mathrm{Nd}}{^{144}\mathrm{Nd}}\right)_{\!0} + \left(\frac{^{147}\mathrm{Sm}}{^{144}\mathrm{Nd}}\right)_{\!{\rm now}}\,(e^{\lambda t}-1) \;} \] (derivation: \(^{147}\mathrm{Sm}_{0} = {}^{147}\mathrm{Sm}_{\rm now}\,e^{\lambda t}\) and the decayed amount is \(^{147}\mathrm{Sm}_0(1-e^{-\lambda t}) = {}^{147}\mathrm{Sm}_{\rm now}(e^{\lambda t}-1)\).) So the initial (ancient) ratio is \[ \left(\frac{^{143}\mathrm{Nd}}{^{144}\mathrm{Nd}}\right)_{\!0} = \left(\frac{^{143}\mathrm{Nd}}{^{144}\mathrm{Nd}}\right)_{\!{\rm now}} - \left(\frac{^{147}\mathrm{Sm}}{^{144}\mathrm{Nd}}\right)_{\!{\rm now}}\,(e^{\lambda t}-1). \] 
Step 2 — compute \(e^{\lambda t}-1\)
\[ \lambda t = 6.54\times10^{-12}\times 4.6\times10^{9} = 0.030084. \] \[ e^{\lambda t}-1 = e^{0.030084}-1 \approx 0.030541\quad(\text{rounded}). \] 
Step 3 — plug numbers in
\[ \Delta = \left(\frac{^{147}\mathrm{Sm}}{^{144}\mathrm{Nd}}\right)_{\!{\rm now}}\,(e^{\lambda t}-1) =0.389\times 0.030541 \approx 0.011884. \] \[ \left(\frac{^{143}\mathrm{Nd}}{^{144}\mathrm{Nd}}\right)_{\!0} =0.516 - 0.011884 \approx 0.504116. \] 
Answer (rounded to three decimals): \[ \boxed{0.504} \]