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標題:
physics:radioactivity
發問:
a rock contained some Th-232 (half life is 1.4 X 10^10 year)at the time it was formed but no lead. The Th-232 decayed and ended up as Pb-208. All steps other than the decay of Th-232 may be regarded as instantaneous. It is found that the ratio of Th-232 atoms to Pb-208 atoms is 6.3. Estimate the age of the... 顯示更多 a rock contained some Th-232 (half life is 1.4 X 10^10 year)at the time it was formed but no lead. The Th-232 decayed and ended up as Pb-208. All steps other than the decay of Th-232 may be regarded as instantaneous. It is found that the ratio of Th-232 atoms to Pb-208 atoms is 6.3. Estimate the age of the rock. THANKS !=( 更新: 唔好意思啊唔係好明呢度: e-kt/(1- e-kt) = 6.3 同埋,, 咁樣又岩唔岩啊?? Use In2/k = half life to find k first Use [A] / [A] initial = e^ -kt to find t [A] / [A] initial = 1/6.3 thx
最佳解答:
From the equation of radioactivity: dN/dt = - kN where k is a constant We have: N = No e-kt where No is the initial no. of radioactive atoms With half-year = 1.4 x 1010 years, we have: (1.4 x 1010)k = ln 2 k = ln 2/(1.4 x 1010) So when ratio of Th-232 : Pb-208 is 6.3 : 1, we have: e-kt/(1- e-kt) = 6.3 e-kt = 6.3 - 6.3 e-kt 7.3 e-kt = 6.3 - kt = ln(6.3/7.3) = - 0.147 t = 0.147(1.4 x 1010)/(ln 2) = 2.976 x 109 years
physics:radioactivity
發問:
a rock contained some Th-232 (half life is 1.4 X 10^10 year)at the time it was formed but no lead. The Th-232 decayed and ended up as Pb-208. All steps other than the decay of Th-232 may be regarded as instantaneous. It is found that the ratio of Th-232 atoms to Pb-208 atoms is 6.3. Estimate the age of the... 顯示更多 a rock contained some Th-232 (half life is 1.4 X 10^10 year)at the time it was formed but no lead. The Th-232 decayed and ended up as Pb-208. All steps other than the decay of Th-232 may be regarded as instantaneous. It is found that the ratio of Th-232 atoms to Pb-208 atoms is 6.3. Estimate the age of the rock. THANKS !=( 更新: 唔好意思啊唔係好明呢度: e-kt/(1- e-kt) = 6.3 同埋,, 咁樣又岩唔岩啊?? Use In2/k = half life to find k first Use [A] / [A] initial = e^ -kt to find t [A] / [A] initial = 1/6.3 thx
最佳解答:
From the equation of radioactivity: dN/dt = - kN where k is a constant We have: N = No e-kt where No is the initial no. of radioactive atoms With half-year = 1.4 x 1010 years, we have: (1.4 x 1010)k = ln 2 k = ln 2/(1.4 x 1010) So when ratio of Th-232 : Pb-208 is 6.3 : 1, we have: e-kt/(1- e-kt) = 6.3 e-kt = 6.3 - 6.3 e-kt 7.3 e-kt = 6.3 - kt = ln(6.3/7.3) = - 0.147 t = 0.147(1.4 x 1010)/(ln 2) = 2.976 x 109 years
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