Problem 20 - Thomas Calculus

Question

Californium- 252 What costs \(\$ 27\) million per gram and can be used to treat brain cancer, analyze coal for its sulfur content, and detect explosives in luggage? The answer is californium- \(252,\) a radioactive isotope so rare that only 8 \(\mathrm{g}\) of it have been made in the western world since its discovery by Glenn Seaborg in 1950 .The half-life of the isotope is 2.645 years - long enough for a useful service life and short enough to have a high radioactivity per unit mass. One microgram of the isotope releases 170 million neutrons per second. a. What is the value of \(k\) in the decay equation for this isotope? b. What is the isotope's mean life? (See Exercise \(19 . )\) c. How long will it take 95\(\%\) of a sample's radioactive nuclei to disintegrate?

Step-by-Step Solution

Answer

a) \( k \approx 0.2615 \text{ year}^{-1} \); b) Mean life is \( 3.825 \) years; c) 95% disintegration takes \( 11.505 \) years.

Step 1: Understand the Decay Equation and Half-life

The formula to calculate the decay constant \( k \) from the half-life \( T_{1/2} \) is \( k = \frac{\ln(2)}{T_{1/2}} \). We know the half-life of Californium-252 is 2.645 years.

Step 2: Calculate Decay Constant \( k \)

Use the formula from Step 1, where \( T_{1/2} = 2.645 \) years. Thus, \( k = \frac{\ln(2)}{2.645} \approx 0.2615 \text{ year}^{-1} \).

Step 3: Understand Mean Life Formula

The mean life \( \tau \) of a radioactive substance is related to the decay constant by the equation \( \tau = \frac{1}{k} \).

Step 4: Calculate the Isotope's Mean Life

Use the formula from Step 3, with \( k = 0.2615 \text{ year}^{-1} \). Thus, \( \tau = \frac{1}{0.2615} \approx 3.825 \) years.

Step 5: Understand the 95% Disintegration Time Formula

To find the time \( t \) for 95% disintegration, use the relation \( N(t) = N_0 e^{-kt} \) where \( N(t) = 0.05 N_0 \). Solve for \( t \) when 95% of the sample has disintegrated.

Step 6: Calculate the Time for 95% Disintegration

We have \( 0.05 = e^{-0.2615t} \). Taking \( \ln \) of both sides gives \( \ln(0.05) = -0.2615t \). Solving for \( t \), we get \( t = \frac{\ln(0.05)}{-0.2615} \approx 11.505 \) years.

Key Concepts

Californium-252

Californium-252 ( ^{252}_{98}Cf) is a highly valued radioactive isotope that is known for its unique applications across various fields. It costs a staggering $27 million per gram, highlighting its rarity and significance. Since its discovery by Glenn Seaborg in 1950, only 8 grams have been produced in the western world. This isotope plays a significant role in nuclear medicine by treating brain cancer thanks to its neutron emission properties.
It is also used in analyzing coal to measure sulfur content and enhancing security by detecting explosives in luggage. One microgram of Californium-252 releases an astounding seventy million neutrons per second, making it extremely potent.

Half-life

The concept of half-life is crucial to understanding radioactive decay. An isotope's half-life is the time required for one-half of the atoms in a radioactive element to decay. For Californium-252, its half-life is 2.645 years. This means that if you start with a certain amount of the isotope, only half of that amount will remain after 2.645 years.
The relatively short half-life of Californium-252 ensures it has a high level of radioactivity, which is beneficial for its applications in medicine and industry. A significant half-life implies the isotope is effective over a useful timeframe for practical uses.

Decay Constant

The decay constant, denoted by the symbol \( k \), is a probability factor that describes the likelihood of decay per unit time for a radioactive isotope. It's a crucial component of the radioactive decay equation, which predicts how much of a sample will remain over time.The relationship between half-life and the decay constant is given by the formula \( k = \frac{\ln(2)}{T_{1/2}} \). For Californium-252, given its half-life of 2.645 years, its decay constant \( k \) is approximately 0.2615 \, \text{year}^{-1}. This figure indicates the highly active nature of Californium-252 and helps in predicting its decay over periods.

Mean Life

Mean life, symbolized as \( \tau \), is another important time measurement in the study of radioactive decay. It represents the average time a radioactive particle exists before decaying. The mean life is longer than the half-life because it considers the decay of all particles in a sample.The connection between mean life and the decay constant is expressed by \( \tau = \frac{1}{k} \). For Californium-252, with its observed decay constant \( k = 0.2615 \text{ year}^{-1} \), the mean life \( \tau \) is approximately 3.825 years. Mean life provides a comprehensive understanding of the isotope's expected longevity.

Nuclear Physics

Nuclear physics is the branch of science that studies the constituents and interactions of atomic nuclei. Californium-252 exemplifies the study of nuclear physics, particularly in how radioactive isotopes decay and the applications of their decay products. Understanding isotopes like Californium-252 requires one to delve into nuclear reactions and radioactive decay processes. This field of physics has profound implications, from energy production to medical treatments, showcasing how fundamental principles of nuclear physics translate to practical solutions in technology and healthcare. Through studying isotopes' decay behaviors, nuclear physics allows us to harness the energy released for a multitude of beneficial applications.