Question

1. the half-life for the radioactive decay of u - 238 is 4.5 billion years and is independent of initial concentration. how long will it take for 10% of the u - 238 atoms in a sample of u - 238 to decay? if a sample of u - 238 initially contained $1.5\times10^{18}$ atoms and was formed 3.8 billion years ago, how many u - 238 atoms does it contain today?

Explanation:

Step1: Find the decay constant

The formula for half - life \(t_{1/2}\) is \(t_{1/2}=\frac{\ln 2}{\lambda}\), where \(\lambda\) is the decay constant. Given \(t_{1/2} = 4.5\times10^{9}\) years, we can solve for \(\lambda\):
\(\lambda=\frac{\ln 2}{t_{1/2}}=\frac{\ln 2}{4.5\times 10^{9}}\text{ years}^{-1}\approx1.54\times 10^{-10}\text{ years}^{-1}\)

Step2: Calculate the time for 10% decay

The radioactive decay formula is \(N = N_{0}e^{-\lambda t}\). If 10% of the atoms decay, then \(N=(1 - 0.1)N_{0}=0.9N_{0}\).
Substitute into the formula: \(0.9N_{0}=N_{0}e^{-\lambda t}\), cancel \(N_{0}\) (since \(N_{0}
eq0\)), we get \(0.9 = e^{-\lambda t}\).
Take the natural logarithm of both sides: \(\ln(0.9)=-\lambda t\), then \(t =-\frac{\ln(0.9)}{\lambda}\)
Substitute \(\lambda=\frac{\ln 2}{4.5\times 10^{9}}\):
\(t=-\frac{\ln(0.9)\times4.5\times 10^{9}}{\ln 2}\)
\(\ln(0.9)\approx - 0.105\), \(\ln 2\approx0.693\)
\(t=\frac{0.105\times4.5\times 10^{9}}{0.693}\approx6.8\times 10^{8}\) years

Step3: Calculate the number of atoms today

Given \(N_{0}=1.5\times 10^{18}\) atoms, \(t = 3.8\times 10^{9}\) years, \(\lambda=\frac{\ln 2}{4.5\times 10^{9}}\)
Use the formula \(N = N_{0}e^{-\lambda t}\)
\(N=1.5\times 10^{18}\times e^{-\frac{\ln 2}{4.5\times 10^{9}}\times3.8\times 10^{9}}\)
\(N = 1.5\times 10^{18}\times e^{-\frac{3.8}{4.5}\ln 2}\)
\(e^{-\frac{3.8}{4.5}\ln 2}=2^{-\frac{3.8}{4.5}}\approx2^{- 0.844}\approx0.57\)
\(N=1.5\times 10^{18}\times0.57 = 8.6\times 10^{17}\) atoms

Answer:

It will take approximately \(6.8\times 10^{8}\) years for 10% of the U - 238 atoms to decay. The sample contains approximately \(8.6\times 10^{17}\) U - 238 atoms today.