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Chapter 10: Problem 33

A sample of radioactive material is obtained from a very old rock. A plot \(\ln A\) verses \(t\) yields a slope value of \(-10^{-9} \mathrm{s}^{-1}\) (see Figure \(10.10(\mathrm{b})\) ). What is the half-life of this material?

Short Answer

Expert verified
The half-life \(T_{1/2}\) of the radioactive material can be found using the decay constant \(\lambda = 10^{-9} \mathrm{s}^{-1}\) and the relation \(T_{1/2} = \frac{\ln 2}{\lambda}\). Substituting the value of \(\lambda\) into the equation, we find that the half-life of this material is approximately \(6.93 \times 10^8\) seconds.

Step by step solution

01

Write down the decay formula

The decay formula is given by: \[A = A_0 e^{-\lambda t}\] Where \(A\) is the activity at time \(t\), \(A_0\) is the initial activity, \(\lambda\) is the decay constant, and \(t\) is the time.
02

Find the decay constant

We are given the slope as \(-10^{-9} \mathrm{s}^{-1}\). Since we have the plot of \(\ln A\) versus \(t\), we can write down the equation for the straight line: \[\ln A = - \lambda t + \ln A_0\] Comparing this with the general form of a straight line, which is \(y = mx + b\) (where \(m\) is the slope and \(b\) is the intercept), we can see that the decay constant \(\lambda\) is equal to the slope (\(-10^{-9} \mathrm{s}^{-1}\)), so: \[\lambda = 10^{-9} \mathrm{s}^{-1}\]
03

Determine the half-life

The half-life of the radioactive material, denoted as \(T_{1/2}\), can be found using the decay constant by the following relation: \[T_{1/2} = \frac{\ln 2}{\lambda}\] Now, we can substitute the value of \(\lambda\) into the equation: \[T_{1/2} = \frac{\ln 2}{10^{-9} \mathrm{s}^{-1}}\]
04

Calculate the half-life

Finally, we can calculate the half-life: \[T_{1/2} = \frac{\ln 2}{10^{-9} \mathrm{s}^{-1}} \approx 6.93 \times 10^8 \mathrm{s}\] So, the half-life of this radioactive material is approximately \(6.93 \times 10^8\) seconds.

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Most popular questions from this chapter

A radioactive sample initially contains \(2.40 \times 10^{-2}\) mol of a radioactive material whose half-life is 6.00 h. How many moles of the radioactive material remain after \(6.00 \mathrm{h}\) ? After 12.0 h? After 36.0 h?

Why does a chain reaction occur during a fission reaction?

The detail that you can observe using a probe is limited by its wavelength. Calculate the energy of a particle that has a wavelength of \(1 \times 10^{-16} \mathrm{m},\) small enough to detect details about one-tenth the size of a nucleon.

(a) Calculate the energy released in the neutron-induced fission reaction \(n+^{235} \mathrm{U} \rightarrow^{92} \mathrm{Kr}+^{142} \mathrm{Ba}+2 n,\) given \(m\left(^{92} \mathrm{Kr}\right)=91.926269 \mathrm{u}\) and \(m\left(^{142} \mathrm{Ba}\right)=141.916361 \mathrm{u} .\) (b) Confirm that the total number of nucleons and total charge are conserved in this reaction.

(a) Calculate the energy released in the \(\alpha\) decay of \(^{238} \mathrm{U} .\) (b) What fraction of the mass of a single \(^{238} \mathrm{U}\) is destroyed in the decay? The mass of \(^{234} \mathrm{Th}\) is 234.043593 u. (c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?
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