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Calculate the mass of a sample of (initially pure) \({ }^{40} \mathrm{~K}\) that has an initial decay rate of \(1.70 \times 10^{5}\) disintegrations/s. The isotope has a half-life of \(1.28 \times 10^{9} \mathrm{y}\).

Short Answer

Expert verified
The mass of the sample is approximately 8.28 grams.

Step by step solution

01

Understand the Relationship Between Decay Rate and Number of Atoms

We begin with the understanding that the decay rate (disintegrations per second) of a radioactive sample is proportional to the number of radioactive atoms present. Mathematically, this is represented as: \(R = \lambda N\), where \(R\) is the decay rate, \(\lambda\) is the decay constant, and \(N\) is the number of atoms. Our goal is to find \(N\) first.
02

Calculate the Decay Constant

The decay constant \(\lambda\) can be calculated using the relation between half-life \(T_{1/2}\) and the decay constant: \(\lambda = \frac{\ln(2)}{T_{1/2}}\). Given \(T_{1/2} = 1.28 \times 10^{9} \text{ years}\) and converting this to seconds: \(1.28 \times 10^{9} \times 3.156 \times 10^{7} \text{ s/year}\), calculate \(\lambda\) using \(1.28 \times 10^{9} \times 3.156 \times 10^{7}\).
03

Calculate Number of Atoms \(N\)

Rearrange the decay equation to solve for \(N\): \(N = \frac{R}{\lambda}\). Substitute the given decay rate \(R = 1.70 \times 10^{5} \text{ s}^{-1}\) and calculated \(\lambda\) to find \(N\).
04

Determine Mass of the Sample

The number of atoms \(N\) relates to the mass \(m\) of the sample using the molar mass \(M\) and Avogadro's number \(N_A\), through the relation \(m = \frac{N \cdot M}{N_A}\). The molar mass of \(^{40}K\) is approximately \(40 \text{ g/mol}\), and Avogadro's number \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\). Substitute \(N\) and solve for \(m\).

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Key Concepts

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

Decay Constant
In the world of radioactive decay, the decay constant (\( \lambda \)) is a very important term. It connects the rate of decay to the actual quantity of radioactive atoms in a sample. This constant is intrinsic to each particular isotope and indicates how fast a sample will decay over time.
To calculate the decay constant, we use the formula: \[\lambda = \frac{\ln(2)}{T_{1/2}}\] where \( \ln(2) \approx 0.693 \) is the natural logarithm of 2, and \( T_{1/2} \) is the half-life of the isotope.
This constant gives us insight into the stability of the isotope—the larger the \( \lambda \), the faster the decay process. In this context, the decay constant exemplifies the exponential nature of decay.
Half-Life
The half-life (\( T_{1/2} \)) of a radioactive substance is the time required for half of the radioactive atoms in a sample to decay into another form. It is a constant value, unique for each radioactive isotope.
When dealing with calculations involving radioactive decay, half-life is essential for determining both the decay constant and understanding how long a sample will remain active.
Understanding half-life allows you to predict the behavior of a decaying isotope over time, providing a measure of its longevity.
It helps in practical applications, such as radioactive dating and medical treatments, where precise decay measurements are critical.
Avogadro's Number
Avogadro's number (\( N_A \)) is a fundamental constant representing the number of atoms, ions, or molecules in one mole of a substance. It's practically always written as \( 6.022 \times 10^{23} \text{ mol}^{-1} \).
This huge number helps us bridge the gap between the microscopic atomic world and our macroscopic everyday world. In radioactive decay, knowing the number of particles we have in a mole allows conversion from number of atoms in a sample to a measurable mass. This is done using the equation: \[m = \frac{N \cdot M}{N_A}\] where \( M \) is the molar mass.
Avogadro's number is thus a crucial factor for converting our calculated number of atoms to a physical, tangible amount.
Molar Mass
Molar mass (\( M \)) is an important concept that relates the number of particles to mass. It is the mass in grams of one mole of atoms, molecules, or ions of a substance.
For example, the molar mass of \(^{40}K\) is approximately 40 grams per mole, which makes calculations involving radioactive decay simpler.
In this context, molar mass allows us to convert from a mathematically calculated number of atoms of \(^{40}K\) to a physical amount we can measure with scales:
\[m = \frac{N \cdot M}{N_A}\] This relationship helps us bridge the theoretical findings of our calculations with practical results pertinent for experimental and real-world use cases.

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Most popular questions from this chapter

A penny has a mass of 3.0 g. Calculate the energy that would be required to separate all the neutrons and protons in this coin from one another. For simplicity, assume that the penny is made entirely of \({ }^{63} \mathrm{Cu}\) atoms (of mass \(62.92960 \mathrm{u}\) ). The masses of the proton-plus-electron and the neutron are \(1.00783 \mathrm{u}\) and \(1.00866 \mathrm{u}\), respectively.

The plutonium isotope \({ }^{239} \mathrm{Pu}\) is produced as a by-product in nuclear reactors and hence is accumulating in our environment. It is radioactive, decaying with a half-life of \(2.41 \times 10^{4} \mathrm{y}\). (a) How many nuclei of Pu constitute a chemically lethal dose of \(2.00 \mathrm{mg} ?\) (b) What is the decay rate of this amount?

A \(5.00 \mathrm{~g}\) charcoal sample from an ancient fire pit has a \({ }^{14} \mathrm{C}\) activity of 63.0 disintegrations/min. A living tree has a \({ }^{14} \mathrm{C}\) activity of 15.3 disintegrations/min per \(1.00 \mathrm{~g}\). The half-life of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{y}\). How old is the charcoal sample?

In \(1992,\) Swiss police arrested two men who were attempting to smuggle osmium out of Eastern Europe for a clandestine sale. However, by error, the smugglers had picked up \({ }^{137} \mathrm{Cs}\). Reportedly, each smuggler was carrying a \(1.0 \mathrm{~g}\) sample of \({ }^{137} \mathrm{Cs}\) in a pocket! In (a) bequerels and (b) curies, what was the activity of each sample? The isotope \({ }^{137} \mathrm{Cs}\) has a half-life of \(30.2 \mathrm{y}\). (The activities of radioisotopes commonly used in hospitals range up to a few millicuries.)

The nuclide \({ }^{198} \mathrm{Au},\) with a half-life of \(2.70 \mathrm{~d},\) is used in cancer therapy. What mass of this nuclide is required to produce an activity of \(250 \mathrm{Ci} ?\)

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