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Chapter 25: Problem 84

Given a radioactive nuclide with \(t_{1 / 2}=1.00 \mathrm{h}\) and a current disintegration rate of 1000 atoms \(s^{-1}\), three hours from now the disintegration rate will be (a) 1000 atoms \(s^{-1} ;\) (b) 333 atoms \(s^{-1} ;\) (c) 250 atoms \(s^{-1}\); (d) 125 atoms \(s^{-1}\)

Short Answer

Expert verified
The disintegration rate for the given radioactive nuclide three hours from now will be 125 atoms \(s^{-1}\), hence option (d) is the correct answer.

Step by step solution

01

Understand the half-life concept

Half-life (\(t_{1 / 2}\)) is the time required for a quantity to reduce to half of its initial value. This is a specific case of a more general concept called exponential decay. The half-life of a substance undergoing decay is constant, not dependent on its amount. In this problem, it's given that the half-life (\(t_{1 / 2}\)) is 1 hour. This implies that the quantity of the radioactive nuclide will halve every hour.
02

Apply the half-life principle to the time specified

As per the half-life principle, after 1 hour (1 half-life), the disintegration rate will reduce to half of its initial rate, i.e., from 1000 atoms \(s^{-1}\) to 500 atoms \(s^{-1}\). Similarly, after 2 hours (2 half-lives), the rate will further reduce to its half, i.e., from 500 atoms \(s^{-1}\) to 250 atoms \(s^{-1}\). Finally, after 3 hours (3 half-lives), the rate will again reduce to its half, i.e., from 250 atoms \(s^{-1}\) to 125 atoms \(s^{-1}\).
03

Compare the result with the given options

Comparing the final disintegration rate (125 atoms \(s^{-1}\)) calculated in step 2 with the given options, it is observed that it matches with option (d).

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Most popular questions from this chapter

Describe how you might use radioactive materials to find a leak in the \(\mathrm{H}_{2}(\mathrm{g})\) supply line in an ammonia synthesis plant.

Explain the important distinctions between each pair of terms: (a) electron and positron; (b) half-life and decay constant; (c) mass defect and nuclear binding energy; (d) nuclear fission and nuclear fusion; (e) primary and secondary ionization.

Radioactive decay and mass spectrometry are often used to date rocks after they have cooled from a magma. \(^{87} \mathrm{Rb}\) has a half-life of \(4.8 \times 10^{10}\) years and follows the radioactive decay $$^{87} \mathrm{Rb} \longrightarrow^{87} \mathrm{Sr}+\beta^{-}$$ A rock was dated by assaying the product of this decay. The mass spectrum of a homogenized sample of rock showed the \(^{87} \mathrm{Sr} /^{86} \mathrm{Sr}\) ratio to be \(2.25 .\) Assume that the original \(^{87} \mathrm{Sr} /^{86} \mathrm{Sr}\) ratio was 0.700 when the rock cooled. Chemical analysis of the rock gave \(15.5 \mathrm{ppm}\) Sr and 265.4 ppm \(\mathrm{Rb},\) using the average atomic masses from a periodic table. The other isotope ratios were \(^{86} \mathrm{Sr} /^{88} \mathrm{Sr}=\) 0.119 and \(^{84} \mathrm{Sr} /^{88} \mathrm{Sr}=0.007 .\) The isotopic ratio for \(^{87} \mathrm{Rb} /^{85} \mathrm{Rb}\) is 0.330. The isotopic masses are as follows:Calculate the following: (a) the average atomic mass of Sr in the rock (b) the original concentration of \(\mathrm{Rb}\) in the rock in \(\mathrm{ppm}\) (c) the percentage of rubidium- 87 decayed in the rock (d) the time since the rock cooled.

The carbon-14 dating method is based on the assumption that the rate of production of \(^{14} \mathrm{C}\) by cosmic ray bombardment has remained constant for thousands of years and that the ratio of \(^{14} \mathrm{C}\) to \(^{12} \mathrm{C}\) has also remained constant. Can you think of any effects of human activities that could invalidate this assumption in the future?

Neutron bombardment of \(^{23}\) Na produces an isotope that is a \(\beta\) emitter. After \(\beta\) emission, the final product is (a) \(^{24} \mathrm{Na} ;\) (b) \(^{23} \mathrm{Mg} ;\) (c) \(^{23} \mathrm{Ar} ;\) (d) \(^{24} \mathrm{Ar} ;\) (e) none of these.
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