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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.

Short Answer

Expert verified
The half-life found in Statement a is not equal to the given half-life of 5.0 minutes. Statement b and Statement c are true, while Statement d is false.

Step by step solution

01

Statement a

: Given that a radioactive nuclide decays from \(1.00 \times 10^{10}\) atoms to \(2.5 \times 10^{9}\) atoms in 10 minutes. Let's calculate the half-life (\(t_{1/2}\)) using the decay equation and given information: Decay equation: \(N = N_0 e^{-\lambda t}\), where \(N\) is the final number of atoms, \(N_0\) is the initial number of atoms, \(\lambda\) is the decay constant, and \(t\) is time elapsed. First, find the decay constant using the given number of initial and final atoms and time elapsed. Rearranging the equation to find the decay constant: \(\lambda = \frac{\ln(N_0/N)}{t}\). Now, substituting given values: \(N_0 = 1.00\times10^{10}\) atoms, \(N = 2.5\times10^9\) atoms, and \(t = 10 \: \text{minutes}\). Calculating the decay constant: \(\lambda = \frac{\ln(1.00\times10^{10}/2.5\times10^9)}{10}\). Now, we will use the relation between the decay constant and half-life: \(t_{1/2} = \frac{\ln(2)}{\lambda}\). Finally, calculating the half-life: \(t_{1/2} = \frac{\ln(2)}{\lambda}\). Compare this value with the given half-life of 5.0 minutes.
02

Statement b

: Large \(Z\) values indicate that the element has a greater number of protons. Elements with large \(Z\) values are more likely to emit \(\alpha\)-particles to reduce the Coulomb repulsion between protons and become more stable. Thus, this statement is true.
03

Statement c

: As \(Z\) increases, the number of protons increases. Due to electrostatic repulsion, the nucleus becomes unstable. Hence, a greater proton-to-neutron ratio is required for stability as observing a higher number of neutrons reduces electrostatic repulsion and increases nuclear attraction. Thus, this statement is true.
04

Statement d

: For light nuclides (less number of nucleons), the most stable isotopes have a proton-to-neutron ratio close to 1:1. If a light nuclide has twice as many neutrons as protons, the proton-to-neutron ratio will be 1:2. This nuclide is not expected to be stable because of the imbalance between protons and neutrons. Thus, this statement is false.

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Key Concepts

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

Half-life Calculation
Radioactive decay is a fascinating process where unstable atomic nuclei lose energy by emitting radiation. A key concept in understanding this process is "half-life," which is the time it takes for half of the radioactive atoms to decay. Imagine pouring a cup of sand and marking when half of it has slipped away—this time is akin to half-life in radioactive decay.

In the example above, to find the half-life, we use the decay equation \(N = N_0 e^{-\lambda t}\), where \(N\) is the remaining number of atoms, \(N_0\) is the initial number, and \(\lambda\) is the decay constant. First, we determine \(\lambda\) using the equation:
  • \(\lambda = \frac{\ln(N_0/N)}{t}\)
After calculating the decay constant, the relation between half-life and decay constant is then expressed as:
  • \(t_{1/2} = \frac{\ln(2)}{\lambda}\)
In our case, the calculation showed a half-life of 5.0 minutes for the decay from \(1.00 \times 10^{10}\) to \(2.5 \times 10^9\) atoms in 10 minutes. If this matches an observed half-life, it depends on accurate calculations and context.
Stability of Nuclides
The stability of nuclides, or atomic nuclei, is determined by the forces holding them together. A stable nuclide remains unchanged over time, while an unstable one undergoes radioactive decay.

Heavy elements with large atomic numbers (high \(Z\) values) often face instability because having too many protons increases electrostatic repulsion. To counterbalance this, such elements frequently emit \(\alpha\)-particles—helium nuclei composed of two protons and two neutrons. This emission reduces the atomic number, increasing stability.

In the context of stability:
  • Heavy nuclides tend to eject \(\alpha\)-particles to lower Coulomb repulsion.
  • The balance between protons and neutrons is a crucial parameter for stability against nuclear forces.
In conclusion, analyzing these atomic interactions helps explain why some nuclides are stable while others are not.
Alpha-particle Emission
Alpha-particle emission is a common type of radioactive decay seen in heavy nuclei. These nuclei emit an \(\alpha\)-particle to achieve a more favorable balance of forces within their core.

What exactly is an \(\alpha\)-particle? Simply put, it is a cluster consisting of two protons and two neutrons—essentially a helium nucleus. This emission reduces the mass number by four units, and the atomic number by two units, thereby transforming the original atom into a different element.

If you observe nuclides with large \(Z\) values that produce \(\alpha\)-particles, this transformative release helps mitigate heavy nucleus instability by:
  • Lowering electrostatic repulsions by reducing proton count.
  • Decreasing the overall mass, aiding in nuclear reconfiguration for stability.
Thus, \(\alpha\)-particle emission is vital in the process of decay for heavy elements, contributing to their path toward achieving nuclear stability.
Proton-to-Neutron Ratio
The proton-to-neutron ratio is essential in determining the stability of a nuclide. It's like balancing a seesaw where protons and neutrons need to find harmony to maintain nuclear stability.

For lighter isotopes, a 1:1 proton-to-neutron ratio often indicates stability. As nuclei grow larger with increasing atomic numbers, the repulsive forces among protons demand more neutrons to stabilize the nucleus.
  • Adding more neutrons offsets proton repulsion by increasing nuclear binding amidst strong nuclear forces.
  • An ideal balance leads to a stable atomic nucleus devoid of spontaneous decay.
Light nuclides usually become unstable when neutrons are disproportionately higher than protons, observed as a 1:2 proton-to-neutron ratio—indicative of predicted instability. Thus, maintaining an appropriate balance within the ratio helps in assessing and understanding nuclear stability.

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Most popular questions from this chapter

A proposed system for storing nuclear wastes involves storing the radioactive material in caves or deep mine shafts. One of the most toxic nuclides that must be disposed of is plutonium-239, which is produced in breeder reactors and has a half-life of 24,100 years. A suitable storage place must be geologically stable long enough for the activity of plutonium- 239 to decrease to \(0.1 \%\) of its original value. How long is this for plutonium- \(239 ?\)

Calculate the binding energy per nucleon for \({ }_{1}^{2} \mathrm{H}\) and \({ }_{1}^{3} \mathrm{H}\). The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410 \mathrm{u} ;\) and \({ }_{1}^{3} \mathrm{H}, 3.01605 \mathrm{u}\).

The bromine- 82 nucleus has a half-life of \(1.0 \times 10^{3}\) min. If you wanted \(1.0 \mathrm{~g}{ }^{82} \mathrm{Br}\) and the delivery time was \(3.0\) days, what mass of NaBr should you order (assuming all of the \(\mathrm{Br}\) in the \(\mathrm{NaBr}\) was \({ }^{82} \mathrm{Br}\) )?

In addition to the process described in the text, a second process called the carbon-nitrogen cycle occurs in the sun: $$ \begin{aligned} { }_{1}^{1} \mathrm{H}+{ }_{6}^{12} \mathrm{C} \longrightarrow{ }_{7}^{13} \mathrm{~N}+{ }_{0}^{0} \gamma \\ { }_{7}^{13} \mathrm{~N} & \longrightarrow{ }_{6}^{13} \mathrm{C}+{ }_{+1}^{0} \mathrm{e} \\ { }_{1}^{1} \mathrm{H}+{ }_{6}^{13} \mathrm{C} &{ }_{7}^{14} \mathrm{~N}+{ }_{0}^{0} \gamma \\ { }_{1}^{1} \mathrm{H}+{ }_{7}^{14} \mathrm{~N} \longrightarrow &{ }_{8}^{15} \mathrm{O}+{ }_{0}^{0} \gamma \\ { }_{8}^{15} \mathrm{O} \longrightarrow{ }_{7}^{15} \mathrm{~N}+{ }_{+1}^{0} \mathrm{e} \\ { }_{1}^{1} \mathrm{H}+{ }_{7}^{15} \mathrm{~N} \longrightarrow{ }_{6}^{12} \mathrm{C}+{ }_{2}^{4} \mathrm{He}+{ }_{0}^{0} \gamma \\ \hline \end{aligned} $$ reaction: \(\quad 4{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{+1}^{0} \mathrm{e}\) a. What is the catalyst in this process? b. What nucleons are intermediates? c. How much energy is released per mole of hydrogen nuclei in the overall reaction? (The atomic masses of \({ }_{1} \mathrm{H}\) and \({ }_{2}^{4} \mathrm{He}\) are \(1.00782 \mathrm{u}\) and \(4.00260 \mathrm{u}\), respectively. \()\)

The mass defect for a lithium-6 nucleus is \(-0.03434 \mathrm{~g} / \mathrm{mol}\). Calculate the atomic mass of \({ }^{6} \mathrm{Li}\)

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