/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 A radioactive isotope of mercury... [FREE SOLUTION] | 91Ó°ÊÓ

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A radioactive isotope of mercury, \({ }^{197} \mathrm{Hg}\), decays to gold, \({ }^{197} \mathrm{Au}\), with a disintegration constant of \(0.0108 \mathrm{~h}^{-1}\). (a) Calculate the half-life of the \({ }^{197} \mathrm{Hg}\). What fraction of a sample will remain at the end of (b) three half-lives and (c) \(10.0\) days?

Short Answer

Expert verified
(a) 64.17 hours, (b) 12.5%, (c) 7.46%

Step by step solution

01

Understand the Half-life Formula

The formula for half-life is given by \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the disintegration constant. In this problem, \( k = 0.0108 \text{ h}^{-1} \).
02

Calculate the Half-life

Using the formula \( t_{1/2} = \frac{0.693}{k} \), substitute \( k = 0.0108 \text{ h}^{-1} \). Thus, \( t_{1/2} = \frac{0.693}{0.0108} \approx 64.17 \text{ hours} \).
03

Calculate Remaining Sample after Three Half-lives

The fraction of the sample remaining after \( n \) half-lives is given by \( \frac{1}{2^n} \). For three half-lives, \( n = 3 \), so the fraction remaining is \( \frac{1}{2^3} = \frac{1}{8} \), which is 0.125 or 12.5% of the original sample.
04

Convert Days to Hours for Part (c)

Since the disintegration constant is in terms of hours, convert 10 days into hours. There are 24 hours in a day, so \( 10 \text{ days} \times 24 \text{ h/day} = 240 \text{ hours} \).
05

Calculate Remaining Sample after 240 Hours

Use the decay formula \( N_t = N_0 \cdot e^{-kt} \) to find the fraction remaining. Here \( k = 0.0108 \text{ h}^{-1} \) and \( t = 240 \text{ h} \). So, the fraction \( N_t / N_0 = e^{-0.0108 \times 240} \approx e^{-2.592} \), which calculates to approximately 0.0746 or 7.46% of the original sample.

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Key Concepts

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

half-life calculation
The half-life of a substance is a measure of how quickly it undergoes radioactive decay. It tells us the time it takes for half of a sample of a radioactive element to disintegrate. This concept is crucial in understanding the stability of the substance over time. The half-life formula is given by:
  • \( t_{1/2} = \frac{0.693}{k} \)
where \( t_{1/2} \) is the half-life and \( k \) is the decay constant. In the given exercise, the decay constant, \( k \), is provided as \( 0.0108 \, \mathrm{h}^{-1} \). By substituting this value into the half-life formula, we can calculate the half-life of the radioactive isotope mercury, \( ^{197} \mathrm{Hg} \), as follows:
  • \( t_{1/2} = \frac{0.693}{0.0108} \approx 64.17 \, \mathrm{hours} \)

This means it takes approximately 64.17 hours for half of a given sample of \( ^{197} \mathrm{Hg} \) to decay into another element.
decay constant
The decay constant is a fundamental parameter in radioactive decay calculations. It is denoted by \( k \) and represents the probability per unit time that a single nucleus will decay. The decay constant is unique for each radioactive isotope and provides essential information on how rapidly the isotope will decay.

Given that the decay constant \( k \) for \( ^{197} \mathrm{Hg} \) is \( 0.0108 \, \mathrm{h}^{-1} \), we can deduce that:
  • The higher the decay constant, the faster the substance will decay.
  • A lower decay constant indicates a slower decay process.

The relationship between the decay constant and half-life is inversely proportional. As seen in the half-life formula,
  • \( t_{1/2} = \frac{0.693}{k} \)

As \( k \) increases, \( t_{1/2} \) decreases, indicating quicker decay. Understanding the decay constant allows us to predict how a radioactive isotope behaves over time. This principle is applied throughout fields like geology, archeology, and medicine.
exponential decay
Exponential decay is a mathematical model describing how quantities decrease rapidly at first, then slowly, with time. This model is highly relevant in understanding radioactive decay processes. The main formula for exponential decay in the context of radioactivity is:
  • \( N_t = N_0 \cdot e^{-kt} \)
where:
  • \( N_t \) = the quantity remaining at time \( t \)
  • \( N_0 \) = the initial quantity
  • \( k \) = decay constant
  • \( t \) = time elapsed
In the given exercise, exponential decay helps us determine how much of \( ^{197} \mathrm{Hg} \) remains after a certain period. For instance, after 240 hours (10 days), the remaining fraction of \( ^{197} \mathrm{Hg} \) can be calculated by substituting \( k = 0.0108 \, \mathrm{h}^{-1} \) and \( t = 240 \, \mathrm{h} \) into the formula:
  • \( N_t / N_0 = e^{-0.0108 \times 240} \approx e^{-2.592} \approx 0.0746 \)
This calculation shows that only about 7.46% of the original sample remains after 10 days, illustrating the principle of exponential decay.

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Most popular questions from this chapter

How many years are needed to reduce the activity of \({ }^{14} \mathrm{C}\) to \(0.020\) of its original activity? The half-life of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{y}\).

Plutonium isotope \({ }^{239}\) Pu decays by alpha decay with a halflife of \(24100 \mathrm{y}\). How many milligrams of helium are produced by an initially pure \(12.0 \mathrm{~g}\) sample of \({ }^{239} \mathrm{Pu}\) at the end of \(20000 \mathrm{y}\) ? (Consider only the helium produced directly by the plutonium and not by any by-products of the decay process.)

Because the neutron has no charge, its mass must be found in some way other than by using a mass spectrometer. When a neutron and a proton meet (assume both to be almost stationary), they combine and form a deuteron, emitting a gamma ray whose energy is \(2.2233 \mathrm{MeV} .\) The masses of the proton and the deuteron are \(1.007276467 \mathrm{u}\) and \(2.013553212 \mathrm{u}\), respectively. Find the mass of the neutron from these data.

A \(1.00 \mathrm{~g}\) sample of samarium emits alpha particles at a rate of 120 particles/s. The responsible isotope is \({ }^{147} \mathrm{Sm}\), whose natural abundance in bulk samarium is \(15.0 \%\). Calculate the half-life.

\(.\) co An \(\alpha\) particle \(\left({ }^{4} \mathrm{He}\right.\) nucleus \()\) is to be taken apart in the following steps. Give the energy (work) required for each step: (a) remove a proton, (b) remove a neutron, and (c) separate the remaining proton and neutron. For an \(\alpha\) particle, what are (d) the total binding energy and (e) the binding energy per nucleon? (f) Does either match an answer to (a), (b), or (c)? Here are some atomic masses and the neutron mass. \(\begin{array}{llll}{ }^{4} \mathrm{He} & 4.00260 \mathrm{u} & { }^{2} \mathrm{H} & 2.01410 \mathrm{u} \\ { }^{3} \mathrm{H} & 3.01605 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ \mathrm{n} & 1.00867 \mathrm{u} & & \end{array}\)

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