Problem 86 - Chemistry

Question

Which of the following statement(s) is (are) true? a. A radioactive nuclide that decays from \(2.00 \times 10^{21}\) atoms to \(5.0 \times 10^{20}\) atoms in 16 minutes has a half-life of 8.0 minutes. b. Nuclides with large \(Z\) values are observed to be \(\alpha\) -particle producers. c. As \(Z\) increases, nuclides need a greater proton-to-neutron ratio for stability. d. Those light nuclides that have twice as many neutrons as protons are expected to be stable.

Step-by-Step Solution

Answer

Statements (b) and (c) are true, while statements (a) and (d) are false.

Step 1: Part a: Determine if the given radioactive nuclide has a half-life of 8 minutes.

We are given the initial number of atoms (2.00 x 10^21), the final number of atoms (5.0 x 10^20), and the time elapsed (16 minutes). The formula for the decay of radioactive nuclides is: \[N(t) = N_0 e^{-\lambda t}\] where \(N(t)\) is the number of atoms at time \(t\), \(N_0\) is the initial number of atoms, \(\lambda\) is the decay constant, and \(t\) is the time elapsed. Let us find the value of the decay constant, \(\lambda\). Rearrange the equation and solve for \(\lambda\): \(e^{-\lambda t} = \frac{N(t)}{N_0}\) \(-\lambda t = \ln{\frac{N(t)}{N_0}}\) \(\lambda = -\frac{\ln{\frac{N(t)}{N_0}}}{t}\) Plug given values into the formula: \(\lambda = -\frac{\ln{\frac{5.0 \times 10^{20}}{2.00 \times 10^{21}}}}{16\:minutes}\) \(\lambda \approx 0.035 \: min^{-1}\) Now, we can use the decay constant \(\lambda\) to find the half-life (\(t_{1/2}\)) by using the formula: \(t_{1/2} = \frac{\ln 2}{\lambda}\) \(t_{1/2} = \frac{\ln 2}{0.035\: min^{-1}}\) \(t_{1/2} \approx 19.71 \:minutes\) Since the calculated half-life is not equal to 8.0 minutes, statement (a) is false.

Step 2: Part b: Assess the connection between large \(Z\) values and α-particle production.

Nuclei with large \(Z\) values (meaning a high number of protons) are more likely to emit α-particles to lower their overall size and energy. α-particle emission consists of the loss of 2 protons and 2 neutrons as a helium nucleus, reducing the \(Z\) value of the original nucleus. This is a well-documented phenomenon, so statement (b) is true.

Step 3: Part c: Analyze the relationship between \(Z\) values and the proton-to-neutron ratio for stability.

As the number of protons (\(Z\) value) in a nucleus increases, the Coulomb repulsive force between them increases too. In order to maintain stability against this repulsive force, a greater number of neutrons are required. This implies that a larger proton-to-neutron ratio is needed for stability as \(Z\) increases. Therefore, statement (c) is true.

Step 4: Part d: Assess the stability criteria for light nuclides with twice as many neutrons as protons.

Generally, in light nucleus with Z < 20, stable nuclides tend to closely follow a 1:1 ratio between the number of protons and neutrons. If there are twice as many neutrons as protons in a given nucleus, it would not satisfy this criterion for stability. This excess in neutrons would typically result in instability due to β-decay. Therefore, statement (d) is false. In conclusion: Statements (b) and (c) are true, while statements (a) and (d) are false.

Key Concepts

Radioactive Decay

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. This decay can occur in several forms, such as alpha, beta, and gamma decay. In this context, we focus on the fundamental decay law, which is crucial in nuclear chemistry.
Radioactive decay is characterized by an exponential decrease in the number of radioactive nuclei over time. The rate of decay is proportional to the number of remaining unstable nuclei. This can be mathematically expressed by the formula: \[ N(t) = N_0 e^{-\lambda t} \] where \( N(t) \) is the number of remaining nuclei at time \( t \), \( N_0 \) is the initial amount, and \( \lambda \) represents the decay constant.

• Decay constant \( \lambda \) is unique to each radioactive substance, reflecting how quickly it undergoes decay.
• The concept is important because it helps determine how long a substance will remain radioactive.

Knowing how to handle this formula allows scientists and students to predict the behavior of radioactive materials over time, which is essential for applications in medicine, industry, and other fields.

Half-life Calculation

The half-life of a radioactive substance represents the time it takes for half of the material to decay, reducing by 50%. This concept is vital in understanding how long a substance will remain hazardous or active in a particular environment.
To find half-life, the relationship with the decay constant \( \lambda \) can be used effectively: \[ t_{1/2} = \frac{\ln 2}{\lambda} \] This equation indicates that half-life is inversely proportional to the decay constant. A larger \( \lambda \) implies a shorter half-life, meaning the substance decays more rapidly.

• Knowing the half-life aids in controlling exposure to radioactive substances by informing necessary safety measures.
• It helps predict when a given amount of radioactive material will reduce to a safe level or lose its effectiveness.

Understanding half-life is vital for making decisions in fields like waste management and nuclear medicine, where timing of radioactive decay dictates safety and usage.

Stability of Nuclides

Nuclides, which are distinct forms of atoms characterized by the number of protons and neutrons they contain, have varying stabilities influenced by their composition.
Key factors affecting nuclide stability include:

• Proton-to-neutron ratio: Lighter nuclides tend to be more stable with a 1:1 ratio, while heavier nuclides require more neutrons to counteract the repulsive forces between protons.
• Energy binding: Nuclides with higher binding energy per nucleon are generally more stable, as it takes more energy to disassemble them.

As the atomic number \( Z \) increases, so too does the requirement for a higher neutron count to maintain stability. This is due to the strong cooperative forces that stabilize a nucleus against the electromagnetic repulsion that occurs between protons.
Understanding nuclide stability is crucial for applications in nuclear reactors and predicting isotopic abundance in natural conditions.

Alpha-particle Emission

Alpha-particle emission is a type of radioactive decay where an unstable nucleus ejects an alpha particle to become more stable. An alpha particle consists of 2 protons and 2 neutrons (the same as a helium nucleus).
This process reduces the atomic number \( Z \) of the original nucleus by 2 and its mass number by 4. Alpha emission is common in heavy elements such as uranium and radium, which undergo such decay to reduce their size and energy.
Key points about alpha-particle emission include:

• It is a form of decay that helps very large, radioactive atoms become smaller and less energetic.
• Upon emission of an alpha particle, the nucleus becomes a different element, moving two places lower on the periodic table.

Alpha-particle emission is less penetrating than other forms of radiation (like beta or gamma) but is highly damaging to living tissue if the source is ingested or inhaled. Understanding this process is important in areas such as radiation shielding and determining the age of rocks and fossils through radiometric dating.