Chemistry, 26.10.2020 17:00, genyjoannerubiera

2) If a rock has 70.7% Uranium-235 (parent) to 29.3% Lead-207 (daughter) isotopes, what is the age of that rock

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answered: imamnaab46

The age of that rock is approximately 353.3 million years.

Explanation:

According to the statement, 29.3 percent of the original mass of Uranium-235 became Lead-207. We know that decay of isotopes is represented by the following ordinary linear differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$ (Eq. 1)

Where:

$\frac{dm}{dt}$ - Rate of change of mass in time, measured in grams per year.

- Current mass of the isotope, measured in grams.

$\tau$ - Time constant, measured in years.

The solution of this differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (Eq. 2)

Where:

$m_{o}$ - Initial mass of the isotope, measured in grams.

- Time, measured in years.

And we solve the expression for time herein:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

$t = -\tau\cdot \ln \frac{m(t)}{m_{o}}$

Besides, time constant can be calculated as a function of half-life. Please notice that half-life of Uranium-235 is 704 million years. The equation is presented below:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

Where $t_{1/2}$ is the half-life of the isotope, measured in years.

If we know that $\frac{m(t)}{m_{o}} = 0.707$ and $t_{1/2} = 704\times 10^{6}\,yr$ , then:

$\tau = \frac{704\times 10^{6}\,yr}{\ln2}$

$\tau \approx 1.016\times 10^{9}\,yr$

$t = -(1.019\times 10^{9}\,yr)\cdot \ln 0.707$

$t \approx 353.312\times 10^{6}\,yr$

The age of that rock is approximately 353.3 million years.

answered: Guest

Hey how did you manage to type such a big lab report

answered: Guest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.