Problem 78 - Chemistry An Atoms First Approach

Question

Which of the following statement(s) is(are) true? a. A radioactive nuclide that decays from \(1.00 \times 10^{10}\) atoms to \(2.5 \times 10^{9}\) atoms in 10 minutes has a half-life of 5.0 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. Statement A is false as the calculated half-life is not equal to 5.0 minutes, and statement D is false because having twice as many neutrons as protons does not guarantee stability even for light nuclides.

Step 1: Statement A

To verify this statement, we need to calculate the half-life of the nuclide. The relationship between initial number of atoms \(N_0\), number of atoms after time 't' \(N_t\), decay constant \(\lambda\) and time 't' is given by: \( N_t = N_0 \cdot e^{-\lambda t} \) Here, \(N_0 = 1.00 \times 10^{10}\), \(N_t = 2.5 \times 10^{9}\), and \(t = 10\) minutes. Substituting these values and solving for \(\lambda\), we get: \(\lambda = 1/t \cdot ln(N_0/N_t) \) The half-life \(T_{1/2}\) is given by: \( T_{1/2} = ln(2) / \lambda \) If the calculated \(T_{1/2}\) is equal to 5.0 minutes, then the statement is true; otherwise, it is false.

Step 2: Statement B

This statement is generally true. Nuclides with large atomic numbers are frequently unstable and undergo alpha decay. Alpha decay is a common form of radioactive decay in which a nucleus emits an alpha particle, reducing its atomic number by two and atomic mass by four. So, we can consider this statement as true.

Step 3: Statement C

This statement is also true. In general, for stable nuclides, the number of neutrons equals (or is slightly more than) the number of protons for light elements. However, as the atomic number increases (which implies large 'Z' values), nuclides need more neutrons than protons for maintaining stability. This is due to the fact that the repulsive forces between protons (which are all positively charged) need to be balanced by the strong nuclear force provided by the neutrons.

Step 4: Statement D

This statement is false. Although as the atomic number increases, nuclides require more neutrons than protons for maintaining stability, having twice as many neutrons as protons does not guarantee stability, even for light nuclides. It is generally seen that stable nuclides have a neutron-proton ratio close to 1 for light nuclides and somewhat greater than 1 for heavy nuclides.

Key Concepts

Half-life Calculation

Calculating the half-life of a radioactive substance gives us insight into the rate at which it decays. The half-life is the time required for half of the radioactive atoms in a sample to decay. For this, we use the radioactive decay formula:\[ N_t = N_0 \cdot e^{-\lambda t} \]where:

• \(N_0\) is the initial amount of the substance
• \(N_t\) is the remaining amount after time \(t\)
• \(\lambda\) is the decay constant
• \(t\) is the elapsed time

To determine the half-life \(T_{1/2}\), we rearrange the decay formula to solve for \(\lambda\) and then calculate:\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]In the context of the problem, if a substance decreases from \(1.00 \times 10^{10}\) atoms to \(2.5 \times 10^{9}\) atoms over 10 minutes, we substitute the values to find \(\lambda\) and then use it to calculate \(T_{1/2}\). If the result matches the given half-life, the statement is correct.

Nuclide Stability

Nuclide stability is heavily influenced by the forces within the nucleus, signified by the proton-neutron balance. Particularly for heavier elements, stability means having the right amount of neutrons to counteract the repulsive forces between positively charged protons. Over a certain atomic number, too many protons can make the nucleus unstable, and extra neutrons are needed to add strong nuclear forces without adding more charge. For lighter elements, stability often occurs at a neutron-to-proton ratio of close to 1. However, as you move to heavier elements, this ratio increases. Beyond this, naturally radioactive nuclides emit particles to become more stable. This observation forms the crux of understanding radioactive processes like alpha decay, aiding in predictions of which nuclides might be stable or radioactive.

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable nucleus emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons resulting in the nucleus losing these 4 nuclear components:

• The atomic number decreases by 2.
• The atomic mass decreases by 4.

This process helps large atomic nuclei increase stability, particularly those with higher atomic numbers. These heavy elements transfer towards a more stable state by releasing energy and decreasing in size. The process is common in elements like uranium and thorium. Understanding alpha decay is crucial as it helps predict the transformation of radioactive materials and explains the presence of helium in certain decay reactions.

Proton-Neutron Ratio

The correct balance of protons to neutrons determines the stability of a nuclide. To maintain a stable structure, lighter elements generally have a proton-neutron ratio close to 1, as both neutrons and protons provide nuclear forces to hold the nucleus together. For heavier elements, more neutrons than protons are needed for stability. This is because, as the element's atomic number increases, the repulsive electromagnetic forces between protons strengthen, requiring additional neutrons to maintain the strong nuclear force that keeps the nucleus intact. Typically, stable nuclides are found in nature when there is an optimal ratio of protons to neutrons. When the ratio deviates significantly from this balance, the nuclide becomes unstable and likely undergoes decay to return to a more stable state.