Problem 59 - Chemistry and Chemical Reactivity

Question

Phosphorus occurs in nature as a single isotope, phosphorus-31. Neutron irradiation of phosphorus-31 produces phosphorus-32, a \(\beta\) emitter with a half-life of 14.28 days. Assume you have a sample containing phosphorus-32 that has a rate of decay of \(3.2 \times 10^{6} \mathrm{dpm} .\) Draw a graph showing disintegrations per minute as a function of time during a period of 1 year.

Step-by-Step Solution

Answer

Draw an exponential decay graph of phosphorus-32 disintegrations per minute over one year.

Step 1: Understand the Problem

We need to graph the decay of phosphorus-32 over a period of one year. Phosphorus-32 decays by emitting beta particles, and its rate of decay is initially \( 3.2 \times 10^6 \) dpm (disintegrations per minute). The half-life of phosphorus-32 is 14.28 days.

Step 2: Calculate the Decay Constant

The decay constant \( \lambda \) can be calculated using the half-life formula: \[ \lambda = \frac{0.693}{t_{1/2}} \] where \( t_{1/2} = 14.28 \text{ days} \times 1440 \text{ minutes/day} \). Substitute and solve to find \( \lambda \).

Step 3: Calculate Number of Atoms at Initial Time

The initial number of phosphorus-32 atoms, \( N_0 \), is related to the decay rate by \( R = \lambda N_0 \). Given \( R = 3.2 \times 10^6 \) dpm, use the decay constant \( \lambda \) from Step 2 to solve for \( N_0 \).

Step 4: Set Up the Decay Equation

The decay of phosphorus-32 can be modeled by the equation: \[ N(t) = N_0 e^{-\lambda t} \] where \( N(t) \) is the number of atoms remaining at time \( t \).

Step 5: Calculate Disintegrations as a Function of Time

The rate of decay \( R(t) \) at time \( t \) is given by: \[ R(t) = \lambda N(t) = \lambda N_0 e^{-\lambda t} \] Calculate \( R(t) \) at various times (e.g., every 14.28 days or every month) over the period of one year.

Step 6: Plot the Graph

On a graph, plot \( R(t) \) with time on the horizontal axis (0 to 365 days) and disintegrations per minute \( R(t) \) on the vertical axis. The curve should show an exponential decay starting from \( 3.2 \times 10^6 \) dpm and decreasing over time.

Key Concepts

Understanding Half-Life Calculations

The concept of half-life is central to understanding radioactive decay processes. When dealing with radioactive substances, the half-life refers to the time it takes for half of the original amount of a radioactive isotope to decay. In this exercise, phosphorus-32 has a half-life of 14.28 days. This means that every 14.28 days, half of the phosphorus-32 in a sample will disintegrate into another element.
The half-life can be a powerful tool for predicting how long a radioactive substance will remain active. It is calculated using the formula:

• \[ t_{1/2} = \frac{0.693}{\lambda} \]

where \( \lambda \) is the decay constant. In practice, understanding and calculating the half-life allows scientists to determine how long a radioactive sample will take to reduce to a certain level of activity or hazard.

Grasping Exponential Decay

Exponential decay is a process where the rate of decay of a substance is proportional to its current value. As phosphorus-32 decays, it follows this pattern, meaning that the amount of radioactive atoms decreases rapidly at first and then more slowly over time. This decay process is described by the equation:

• \[ N(t) = N_0 e^{-\lambda t} \]

where \( N(t) \) is the amount of substance remaining at time \( t \), \( N_0 \) is the initial amount of the substance, and \( \lambda \) is the decay constant.
The exponential decay nature makes it useful for various applications in science and engineering, especially when predicting how quickly a material will transform or reduce to a safer level. When plotted on a graph, the exponential decay curve starts high and gradually decreases, illustrating how fast the substance loses its radioactive quality over time.

Exploring Beta Radiation

Beta radiation is one type of decay that occurs in radioactive substances. In the case of phosphorus-32, it emits beta particles as it decays. Beta particles are high-energy, high-speed electrons or positrons that are ejected from the nucleus of the decaying atom. This process transforms the phosphorus-32 nucleus into a more stable form, typically sulfur-32.
Understanding beta radiation is important because it has direct implications for safety and usage of radioactive materials. Beta particles can penetrate the skin but are generally stopped by materials like plastic or glass, making them less penetrating than gamma rays, but more so than alpha particles.
Therefore, handling materials that emit beta radiation requires specific safety measures to protect against exposure, such as wearing protective clothing and using shielding materials. Knowing the type of radiation a material emits helps in designing proper storage and disposal methods to minimize risks associated with its use.