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Chapter 23: Problem 51

Some of the reactions explored by Rutherford and others are listed below. Identify the unknown species in each reaction. (a) \(^{14}_{7} \mathrm{p}+_{2}^{4} \mathrm{He} \longrightarrow_{8}^{17} \mathrm{O}+?\) (b) \(_{4}^{9} \mathrm{Be}+_{2}^{4} \mathrm{He} \longrightarrow ?+_{0}^{1} \mathrm{n}\) (c) \(?+_{2}^{4} \mathrm{He} \longrightarrow_{15}^{30} \mathrm{P}+_{0}^{1} \mathrm{n}\) (d) \(^{239}_{94} \mathrm{Pu}+_{2}^{4} \mathrm{He} \longrightarrow ?+_{0}^{1} \mathrm{n}\)

Short Answer

Expert verified
(a) \( ^{1}_{1}\mathrm{p} \), (b) \( _{6}^{12}\mathrm{C} \), (c) \( _{13}^{27}\mathrm{Al} \), (d) \( _{96}^{242}\mathrm{Cm} \).

Step by step solution

01

Reaction (a) - Subtract to find unknown

In the reaction \( ^{14}_{7} \mathrm{N} + _{2}^{4} \mathrm{He} \longrightarrow _{8}^{17} \mathrm{O} + ? \), apply the conservation of mass number and atomic number. The mass number before and after the reaction must be equal as well as the atomic number.**Mass numbers:** \(14 + 4 = 17 + A?\) leading to \(A? = 1\).**Atomic numbers:** \(7 + 2 = 8 + Z?\) leading to \(Z? = 1\).This implies the unknown is a proton (\( ^{1}_{1}\mathrm{p} \)).
02

Reaction (b) - Apply conservation laws

For the reaction \( _{4}^{9} \mathrm{Be} + _{2}^{4} \mathrm{He} \longrightarrow ? + _{0}^{1} \mathrm{n} \), use the conservation laws.**Mass numbers:** \(9 + 4 = A? + 1\) leading to \(A? = 12\).**Atomic numbers:** \(4 + 2 = Z? + 0\) leading to \(Z? = 6\).Thus, the unknown is \( _{6}^{12}\mathrm{C} \) (Carbon).
03

Reaction (c) - Reverse solving for unknown

Given \( ? + _{2}^{4} \mathrm{He} \longrightarrow _{15}^{30} \mathrm{P} + _{0}^{1} \mathrm{n} \), we find the reactant by reversing the conservation equations.**Mass numbers:** \(A? + 4 = 30 + 1\) leading to \(A? = 27\).**Atomic numbers:** \(Z? + 2 = 15 + 0\) leading to \(Z? = 13\).Therefore, the unknown species is \( _{13}^{27}\mathrm{Al} \) (Aluminium).
04

Reaction (d) - Identify unknown species

For \( ^{239}_{94} \mathrm{Pu} + _{2}^{4} \mathrm{He} \longrightarrow ? + _{0}^{1} \mathrm{n} \), follow the conservation.**Mass numbers:** \(239 + 4 = A? + 1\) resulting in \(A? = 242\).**Atomic numbers:** \(94 + 2 = Z? + 0\) leading to \(Z? = 96\).Thus, the unknown is \( _{96}^{242}\mathrm{Cm} \) (Curium).

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Key Concepts

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

Conservation of Mass Number
The conservation of mass number is an essential concept in nuclear reactions. This principle states that the total mass number (the sum of protons and neutrons) must remain the same before and after a reaction. This allows us to balance nuclear equations effectively.
For instance, in the case of reaction (a), where nitrogen-14 is involved with helium-4 to produce oxygen-17, we used this law to find that the missing particle is a proton. We calculated the mass numbers as follows:
  • The sum of the mass numbers on the left side of the equation: 14 (nitrogen) + 4 (helium) = 18.
  • The sum on the right side: 17 (oxygen) + X = 18.
Here, X must be 1, indicating the presence of a proton. This example illustrates the significance of conserving the mass number to identify unknown species in nuclear reactions.
Conservation of Atomic Number
The conservation of atomic number is another vital aspect of nuclear reactions. This rule ensures the total number of protons, represented by the atomic number, remains constant through the reaction. This aids us in predicting and balancing nuclear changes.
In reaction (a), after applying the conservation of atomic number:
  • Adding the atomic numbers on the reactants side: 7 (nitrogen) + 2 (helium) = 9.
  • Which must equal the atomic numbers on the products side: 8 (oxygen) + Z.
  • This reveals that Z is 1.
Thus, we identified the emitted particle as a proton (symbolized as \(^{1}_{1} ext{p} \). This principle confirms that nuclear reactions rearrange but do not create or destroy protons.
Proton Emission
Proton emission occurs when a nucleus releases a proton, resulting in a decrease in its atomic number by one unit while the mass number remains unchanged. This is less common but important in identifying elements and balancing nuclear reactions.
Take reaction (a) as an example again; the identification of the unknown as a proton signifies proton emission. This results in the transformed nucleus (oxygen) having one less proton compared to the combined nuclei before the reaction.
  • This adjustment in atomic structure is crucial for altering the identity of the involved element.
  • Therefore, understanding proton emission helps us predict outcomes and confirm the products formed in nuclear reactions.
Neutron Emission
Neutron emission is a type of nuclear reaction where a neutron is released from the nucleus. This event doesn鈥檛 change the atomic number but reduces the mass number by one.
In step (b), after completing the necessary conservation calculations, we observed this in action. When beryllium-9 and helium-4 react, they produce carbon-12 and emit a neutron.
  • Thus reflecting a decrease in mass number while retaining the same atomic number for the newly formed element.
  • Similarly, steps related to reactions (c) and (d) also show neutron emissions, helping in balancing the nuclear equations and identifying the resultant elements.
Neutron emissions are instrumental in stabilizing the nucleus as well as detailing transformations in nuclear reactions.

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Most popular questions from this chapter

When a neutron is captured by an atomic nucleus, energy is released as \(\gamma\) radiation. This energy can be calculated based on the change in mass in converting reactants to products. For the nuclear reaction \(_{3}^{6} \mathrm{Li}+_{0}^{1} \mathrm{n} \longrightarrow_{3}^{7} \mathrm{Li}+\gamma:\) (a) Calculate the energy evolved in this reaction (per atom). Masses needed for this calculation are \(_{3}^{6} \mathrm{Li}=\) \(6.01512,_{0}^{1} n=1.00867,\) and \(_{3}^{7} \mathrm{Li}=7.01600\). (b) Use the answer in part (a) to calculate the wavelength of the \(\gamma\) -rays emitted in the reaction.

A sample of wood from a Thracian chariot found in an excavation in Bulgaria has a \(^{14} \mathrm{C}\) activity of \(11.2 \mathrm{dpm} / \mathrm{g}\) Estimate the age of the chariot and the year it was made \((t_{1 / 2} \text { for }^{14} \mathrm{C} \text { is } 5.73 \times 10^{3} \text { years, and the activity of }^{14} \mathrm{C}\) in. living material is \(14.0 \mathrm{dpm} / \mathrm{g}).\)

Element \(^{287} 114\) decayed by \(\alpha\) emission with a half-life of about \(5 s\). Write an equation for this process.

Sodium-23 (in a sample of NaCl) is subjected to neutron bombardment in a nuclear reactor to produce \(^{24}\) Na. When removed from the reactor, the sample is radioactive, with \(\beta\) activity of \(2.54 \times 10^{4} \mathrm{dpm} .\) The decrease in radioactivity over time was studied, producing the following data: $$\begin{array}{lc}\hline \text { Activity }(\mathrm{dpm}) & \text { Time }(\mathrm{h}) \\\\\hline 2.54 \times 10^{4} & 0 \\ 2.42 \times 10^{4} & 1 \\\2.31 \times 10^{4} & 2 \\\2.00 \times 10^{4} & 5 \\\1.60 \times 10^{4} & 10 \\\1.01 \times 10^{4} & 20 \\\\\hline\end{array}$$ (a) Write equations for the neutron capture reaction and for the reaction in which the product of this reaction decays by \(\beta\) emission. (b) Determine the half-life of sodium- 24.

The age of minerals can sometimes be determined by measuring the amounts of \(^{206} \mathrm{Pb}\) and \(^{238} \mathrm{U}\) in a sample. This determination assumes that all of the \(^{206} \mathrm{Pb}\) in the sample comes from the decay of \(^{238} \mathrm{U}\). The date obtained identifies when the rock solidified. Assume that the ratio of \(^{206} \mathrm{Pb}\) to \(^{238} \mathrm{U}\) in an igneous rock sample is \(0.33 .\) Calculate the age of the rock. \((t_{1 / 2} \text { for }^{238} \mathrm{U} \text { is } 4.5 \times 10^{9}\) years.).
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