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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

half-life
The concept of half-life is central to understanding radioactive decay. It measures the time it takes for half of a given quantity of a radioactive substance to decay. This property is unique in that it is constant for each radioactive element, irrespective of the initial amount of the substance.

Consider a radioactive nuclide with a half-life of 1 hour. If you begin with a disintegration rate, or rate of radioactive decay, of 1000 atoms per second, the half-life tells us that in one hour, this rate will reduce to half, which is 500 atoms per second.

The calculation of half-life is a repetitive division by two, showing its reliance on time rather than the quantity of substance. Each hour reduces the disintegration rate by half, indicating a predictable exponential decay pattern.
exponential decay
Exponential decay is a process in which the quantity of a substance decreases at a rate proportional to its current value. This kind of decay is common among radioactive substances, including the nuclides studied in physics and chemistry.

The exponential nature of radioactive decay implies that over equal time intervals, the substance reduces by a consistent fraction of its existing amount, generally half. For instance, if the decay follows the exponential pattern with a half-life of 1 hour, it will consistently halve every hour.

Mathematically, this can be represented by the formula: \[ N(t) = N_0 \times \left( \frac{1}{2} \right)^{t/t_{1/2}} \]where \( N(t) \) is the quantity remaining after time \( t \), \( N_0 \) is the initial quantity, and \( t_{1/2} \) is the half-life. This equation best showcases how time governs decay, not the magnitude of the initial amount.
disintegration rate
The disintegration rate is a measure of how quickly atoms in a radioactive sample are decaying. It indicates the number of atoms that disintegrate, or decay, per unit time.

In practical terms, measuring the disintegration rate allows scientists to determine the activity of a radioactive substance. For a nuclide with a half-life of 1 hour and an initial disintegration rate of 1000 atoms per second, the activity decreases as time progresses.

With each passing hour, or each half-life, that rate halves because the remaining quantity of radioactive material continues to decrease, significantly influencing the overall activity. Over three hours, the disintegration rate changes from 1000 atoms per second to 125 atoms per second, following a consistent pattern of exponential decay.

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Most popular questions from this chapter

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.

One member each of the following pairs of radioisotopes decays by \(\beta^{-}\) emission, and the other by positron \(\left(\beta^{+}\right)\) emission: \((\mathrm{a})_{15}^{29} \mathrm{P}\) and \(_{15}^{33} \mathrm{P} ;(\mathrm{b}) \stackrel{120}{53} \mathrm{I}\) and \(_{53}^{134} \mathrm{I} .\) Which is which? Explain your reasoning.

A sample containing \(_{88}^{224} \mathrm{Ra},\) which decays by \(\alpha\) -particle emission, disintegrates at the following rate, expressed as disintegrations per minute or counts per minute \((\mathrm{cpm}): t=0,1000 \mathrm{cpm} ; t=1 \mathrm{h}\) \(992 \mathrm{cpm} ; t=10 \mathrm{h}, 924 \mathrm{cpm} ; t=100 \mathrm{h}, 452 \mathrm{cpm}\) \(t=250 \mathrm{h}, 138 \mathrm{cpm} .\) What is the half-life of this nuclide?

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 most radioactive of the isotopes of an element is the one with the largest value of its (a) half-life, \(t_{1 / 2}\) (b) neutron number, \(N ;\) (c) mass number, \(Z\) (d) radioactive decay constant, \(\lambda\)

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