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A zircon crystal is measured for age using uranium-235 to lead-207 decay (half-life is 704 million years). Assuming that there was no lead207 initially, and you measure the elemental composition and find 35,000,000 lead-207 atoms and 1,100,000 uranium-235 atoms, how old is the zircon crystal?

66. A zircon crystal is measured for age using uranium235-235 to lead207-207 decay (half-life is 704704 million years). Assuming that there was no lead207207 initially, and you measure the elemental composition and find 3535,000000,000000 lead207-207 atoms and 11,100100,000000 uranium235-235 atoms, how old is the zircon crystal?

Full solution

Q. 66. A zircon crystal is measured for age using uranium235-235 to lead207-207 decay (half-life is 704704 million years). Assuming that there was no lead207207 initially, and you measure the elemental composition and find 3535,000000,000000 lead207-207 atoms and 11,100100,000000 uranium235-235 atoms, how old is the zircon crystal?
  1. Calculate Initial Amount: Calculate the initial amount of uranium235-235. Since the half-life of uranium235-235 is 704704 million years, and we know the current amounts of lead207-207 and uranium235-235, we can find the initial amount of uranium235-235. Initial uranium235-235 == current uranium235-235 ++ lead207-207 =1,100,000+35,000,000=36,100,000= 1,100,000 + 35,000,000 = 36,100,000.
  2. Determine Number of Half-Lives: Determine the number of half-lives that have passed.\newlineUsing the formula for decay, N=N0×(1/2)t/T N = N_0 \times (1/2)^{t/T} , where N N is the remaining amount of uranium235-235, N0 N_0 is the initial amount, t t is the time elapsed, and T T is the half-life.\newlineRearranging the formula to solve for t t , we get t=T×log2(N0/N) t = T \times \log_2(N_0/N) .\newlineSubstituting the values, t=704×log2(36,100,000/1,100,000) t = 704 \times \log_2(36,100,000 / 1,100,000) .
  3. Calculate Logarithm and Age: Calculate the logarithm and the age.\newlineCalculating log2(36,100,000/1,100,000) \log_2(36,100,000 / 1,100,000) = log2(32.8181818) \log_2(32.8181818) 55.035035.\newlineThen, t=704×5.035 t = 704 \times 5.035 35443544.6464 million years.

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