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Chapter 42: Problem 29

A radioactive sample intended for irradiation of a hospital patient is prepared at a nearby laboratory. The sample has a halflife of \(83.61 \mathrm{~h}\). What should its initial activity be if its activity is to be \(7.4 \times 10^{8} \mathrm{~Bq}\) when it is used to irradiate the patient \(24 \mathrm{~h}\) later?

Short Answer

Expert verified
The initial activity should be approximately \( 8.95 \times 10^8 \mathrm{~Bq} \).

Step by step solution

01

Understanding Half-Life

The half-life of a radioactive material is the time it takes for half of the sample to decay. In this case, the half-life is given as 83.61 hours. This means that every 83.61 hours, the activity of the sample reduces to half of its initial value.
02

Calculate Decay Constant

The decay constant \( \lambda \) is related to the half-life through the formula \( \lambda = \frac{\ln(2)}{T_{1/2}} \). Substituting the half-life \( 83.61 \mathrm{~h} \) into the formula gives \[ \lambda = \frac{\ln(2)}{83.61} \] which calculates to approximately \( 0.00829 \mathrm{~h}^{-1} \).
03

Initial Activity Formula

The activity of a radioactive sample at a specific time can be expressed by the formula: \( A(t) = A_0 e^{-\lambda t} \), where \( A_0 \) is the initial activity, \( \lambda \) is the decay constant, and \( t \) is the time elapsed. We will use this formula to find the initial activity \( A_0 \).
04

Rearranging to Find Initial Activity

We need to find \( A_0 \) given \( A(24) = 7.4 \times 10^8 \mathrm{~Bq} \) and \( t = 24 \mathrm{~h} \). Rearranging the activity formula gives \[ A_0 = \frac{A(t)}{e^{-\lambda t}} = A(t) \times e^{\lambda t} \].
05

Calculate Exponential Decay Component

Substitute \( \lambda = 0.00829 \mathrm{~h}^{-1} \) and \( t = 24 \mathrm{~h} \) into \( e^{-\lambda t} \). The calculation \( e^{-0.00829 \times 24} \) results in approximately \( 0.827 \).
06

Calculate Initial Activity

Using the rearranged formula \( A_0 = 7.4 \times 10^8 \times \frac{1}{0.827} \), the initial activity \( A_0 \) is then calculated to be approximately \( 8.95 \times 10^8 \mathrm{~Bq} \).

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Key Concepts

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

Half-Life
The half-life of a radioactive material is an essential concept in nuclear physics. It's defined as the time it takes for half of a given sample to decay. Imagine you start with 100 grams of a radioactive material. After one half-life, only 50 grams will remain. This process continues, so after another half-life, you'd have 25 grams.
Half-life helps us predict how long a radioactive sample will remain active. In the example given, the half-life is 83.61 hours. This means that over any 83.61-hour period, the activity of the sample decreases by half.
  • Calculating half-life is crucial when assessing the safety and effectiveness of radioactive samples, especially in medical applications.
  • It allows researchers and medical professionals to determine the timing for using a sample effectively.
Activity of Radioactive Sample
The activity of a radioactive sample measures its rate of decay and is often expressed in becquerels (Bq), where 1 Bq represents one decay per second. In practical terms, activity indicates how "active" a sample is at a given time.
Knowing the activity is crucial for using radioactive materials efficiently, especially in medical treatments where precise doses are required.
  • In our example, the desired activity when the patient is irradiated is 7.4 x 108 Bq.
  • We must calculate the initial activity accurately so that the desired activity level is achieved at the necessary time.
Decay Constant
The decay constant, symbolized by \( \lambda \), is key to understanding how quickly a radioactive sample decays. It's found using the formula \( \lambda = \frac{\ln(2)}{T_{1/2}} \), where \( T_{1/2} \) is the half-life.
In simpler terms, the decay constant is a measure of the probability that a radioactive nucleus will decay per unit time.
  • In our scenario with a half-life of 83.61 hours, \( \lambda \) calculates to approximately 0.00829 per hour.
  • This tells us the rate at which the sample's atoms are expected to transform.
Exponential Decay
Exponential decay describes how the quantity of a radioactive sample decreases over time. The formula \( A(t) = A_0 e^{-\lambda t} \) shows how the activity changes based on time \( t \), the initial activity \( A_0 \), and the decay constant \( \lambda \).
  • The formula is key to finding out what the initial activity should be to ensure the sample's effectiveness at any future point.
  • Exponential decay ensures that the decrease in activity isn't constant but diminishes more slowly over time.
For example, the calculation in the provided exercise shows how using this formula, we can rearrange it to solve for \( A_0 \), giving us the necessary starting activity to reach the desired future level.

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Most popular questions from this chapter

The isotope \({ }^{238} \mathrm{U}\) decays to \({ }^{206} \mathrm{~Pb}\) with a half-life of \(4.47 \times 10^{9} \mathrm{y}\). Although the decay occurs in many individual steps, the first step has by far the longest half-life; therefore, one can often consider the decay to go directly to lead. That is, \({ }^{238} \mathrm{U} \rightarrow{ }^{206} \mathrm{~Pb}+\) various decay products. A rock is found to contain \(4.20 \mathrm{mg}\) of \({ }^{238} \mathrm{U}\) and \(2.135 \mathrm{mg}\) of \({ }^{206} \mathrm{~Pb}\). Assume that the rock contained no lead at formation, so all the lead now present arose from the decay of uranium. How many atoms of (a) \({ }^{238} \mathrm{U}\) and (b) \({ }^{206} \mathrm{~Pb}\) does the rock now contain? (c) How many atoms of \({ }^{238} \mathrm{U}\) did the rock contain at formation? (d) What is the age of the rock?

The half-life of a radioactive isotope is \(120 \mathrm{~d}\). How many days would it take for the decay rate of a sample of this isotope to fall to one- fourth of its initial value?

A certain radionuclide is being manufactured in a cyclotron at a constant rate \(R\). It is also decaying with disintegration constant \(\lambda\). Assume that the production process has been going on for a time that is much longer than the half-life of the radionuclide. (a) Show that the number of radioactive nuclei present after such time remains constant and is given by \(N=R / \lambda\). (b) Now show that this result holds no matter how many radioactive nuclei were present initially. The nuclide is said to be in secular equilibrium with its source; in this state its decay rate is just equal to its production rate.

What is the binding energy per nucleon of the rutherfordium isotope \({ }_{104}^{259} \mathrm{Rf} ?\) Here are some atomic masses and the neutron mass. \(\begin{array}{lccc}{ }_{104}^{259} \mathrm{Rf} & 259.10563 \mathrm{u} & { }^{1} \mathrm{H} & 1.007825 \mathrm{u} \\ \mathrm{n} & 1.008665 \mathrm{u} & & \end{array}\)

Under certain rare circumstances, a nucleus can decay by emitting a particle more massive than an alpha particle. Consider the decays $$ { }^{223} \mathrm{Ra} \rightarrow{ }^{209} \mathrm{~Pb}+{ }^{14} \mathrm{C} \quad \text { and } \quad{ }^{223} \mathrm{Ra} \rightarrow{ }^{219} \mathrm{Rn}+{ }^{4} \mathrm{He} $$ Calculate the \(Q\) value for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is \(30.0 \mathrm{MeV}\). What is the barrier height for \({ }^{14} \mathrm{C}\) emission? (Be careful about the nuclear radii.) The needed atomic masses are \(\begin{array}{lllr}{ }^{223} \mathrm{Ra} & 223.01850 \mathrm{u} & { }^{14} \mathrm{C} & 14.00324 \mathrm{u} \\ { }^{209} \mathrm{~Pb} & 208.98107 \mathrm{u} & { }^{4} \mathrm{He} & 4.00260 \mathrm{u} \\ { }^{219} \mathrm{Rn} & 219.00948 \mathrm{u} & & \end{array}\)
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