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Suppose we bombard nitrogen with alpha particles. What might result if an alpha is absorbed and a proton subsequently is emitted? [Hint: Write out an equation accounting for all the particles.]

Short Answer

Expert verified
The reaction produces oxygen: \( ^{14}_{7}\text{N} + ^{4}_{2}\text{He} \rightarrow ^{17}_{8}\text{O} + ^{1}_{1}\text{p} \).

Step by step solution

01

Identify the Particles Involved

Start by identifying the particles mentioned in the exercise. Nitrogen, an alpha particle, and a proton. A nitrogen atom is normally represented as \( ^{14}_{7}\text{N} \). The alpha particle, which is a helium nucleus, is represented as \( ^{4}_{2}\text{He} \). Finally, a proton is represented as \( ^{1}_{1}\text{p} \).
02

Write the Nuclear Reaction Equation

The problem describes a reaction where an alpha particle is absorbed by nitrogen, and a proton is emitted. The initial particles are \( ^{14}_{7}\text{N} \) and \( ^{4}_{2}\text{He} \), while the product side needs to account for the proton emission, represented as \( ^{1}_{1}\text{p} \). Therefore, the nuclear reaction is:\[ ^{14}_{7}\text{N} + ^{4}_{2}\text{He} \rightarrow X + ^{1}_{1}\text{p} \] where \(X\) is the resultant nucleus.
03

Determine the Resultant Nucleus

To find the resultant nucleus \(X\), balance both atomic and mass numbers between reactants and products. The total mass number initially is \(18\) (\(14 + 4\)) and the atomic number is \(9\) (\(7 + 2\)). After the emission of a proton, the mass number of \(X\) will be \(17\) (\(18 - 1\)) and the atomic number \(8\) (\(9 - 1\)).
04

Identify the Resultant Element

The element with atomic number \(8\) is oxygen (\(^{17}_{8}\text{O}\)). Therefore, the resultant nuclear reaction that occurs is:\[ ^{14}_{7}\text{N} + ^{4}_{2}\text{He} \rightarrow ^{17}_{8}\text{O} + ^{1}_{1}\text{p} \]. The reaction transforms a nitrogen atom into an oxygen atom after absorbing an alpha particle and emitting a proton.

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Key Concepts

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

Alpha Particles
Alpha particles are a fascinating component of nuclear reactions. They are essentially helium nuclei, meaning they consist of 2 protons and 2 neutrons. This gives them a mass number of 4 and an atomic number of 2, represented as \(^{4}_{2}\text{He}\). These particles are relatively heavy and carry a positive charge due to their protons.

In the context of nuclear reactions, alpha particles play a critical role. They can initiate reactions when they collide with other nuclei because of their high energy and mass. This property makes them effective in processes like the one described in the exercise, where an alpha particle is absorbed by nitrogen. The absorption alters the balance of the nucleus, leading to potential transformations, such as the emission of a proton.

It's essential to understand that while alpha particles are a crucial component in nuclear reactions, their interaction with matter is relatively limited, as they cannot penetrate deeply due to their mass and charge. This means they have practical applications in situations where surface-level interactions are sufficient.
Nitrogen
Nitrogen is a vital element on Earth, known for its role in the atmosphere and various biological processes. In nuclear chemistry, nitrogen is denoted as \(^{14}_{7}\text{N}\). The subscript shows its atomic number, 7, which indicates the number of protons, while the superscript 14 represents its mass number, comprising both protons and neutrons.

When nitrogen is involved in nuclear reactions, like in the exercise, it can undergo significant changes. Upon bombardment by alpha particles, nitrogen can absorb the alpha particle, prompting a transformation in its nuclear structure. This absorption can lead to further nuclear adjustments, which might include the emission of different particles—such as a proton.

The study of nitrogen in nuclear reactions gives insight into how elements can transform under specific conditions, revealing the dynamic nature of atomic nuclei. These reactions have various applications, including in medical and energy-generation fields, demonstrating nitrogen's versatility beyond its gaseous form.
Proton Emission
Proton emission is a nuclear process where a proton, the positively charged particle within an atomic nucleus, is ejected. In nuclear equations, a proton is represented as \(^{1}_{1}\text{p}\), reflecting its mass number and atomic number, each being 1. This process can occur spontaneously in certain unstable nuclei or, as in our exercise example, as a result of a preceding reaction like the absorption of an alpha particle.

When nitrogen absorbs an alpha particle, it becomes energetically excited and unstable. To restore stability, the nucleus can emit a proton, leading to a change in both the atomic and mass numbers. This emission results in the nucleus transforming into a different element with one less proton. Hence, nitrogen transformed into oxygen upon the emission of a proton.

Proton emission is a subtle yet significant aspect of nuclear physics. It helps understand changes at an atomic level, offering insights into isotope transformation and energy release during nuclear reactions. Understanding the concept of proton emission is key to comprehending broader nuclear processes such as radioactive decay and artificial transmutation of elements.
Nuclear Equations
Nuclear equations are vital for understanding and representing nuclear reactions like those involving nitrogen and alpha particles. These equations demonstrate the conservation of atomic and mass numbers during nuclear transformations.

The basic format of a nuclear equation shows reactants on the left and products on the right. Arrow separates the two, indicating transformation or process direction. For instance, our exercise equation starts as \(^{14}_{7}\text{N} + ^{4}_{2}\text{He} \rightarrow X + ^{1}_{1}\text{p}\), leading to the identification of oxygen as \(X\), the resultant product.

In writing these equations, it's crucial to balance the atomic numbers (protons) and the mass numbers (total of protons and neutrons). These balances ensure the adherence to conservation laws of mass and energy, affirming the integrity of nuclear reactions.

Grasping the function of nuclear equations unravels the complexities of nuclear transformations. They are indispensable tools that facilitate the learning and application of nuclear chemistry concepts, bridging the connection between theoretical knowledge and practical experimentation.

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Most popular questions from this chapter

What mass of \({ }_{27}^{60} \mathrm{Co}\) has an activity of \(1.0\) Ci? The half-life of cobalt-60 is \(5.25\) years.

A certain isotope has a half-life of \(7.0 \mathrm{~h}\). How many seconds does it take for 10 percent of the sample to decay?

Complete the following nuclear equations: \((a)^{1}\) ic) \(\operatorname{tBe}(p, a) ?\) \(\operatorname{sicat} \alpha, 24\) (a) The sum of the subscripts on the left is \(7+2=9\). The subscript of the first product on the right is 8 . Hence, the second product on the right must have a subscript (net charge) of 1 . Also, the sum of the superscripts on the left is \(14+4=18\). The superscript of the first product is 17 . Hence, the second product on the right must have a superscript (mass number) of 1 . The particle with nuclear charge 1 and mass number 1 is the proton, \({ }_{1}^{1} \mathrm{H} .\) (b) The nuclear charge of the second product particle (its subscript) is \((4+2)-6=0\). The mass number of the particle (its superscript) is \((9+4)-12=1\). Hence, the particle must be the neutron, \({ }_{0}^{1} n\). (c) The reactants \({ }_{4}^{9} \mathrm{Be}\) and \({ }_{1}^{1} \mathrm{H}\) have a combined nuclear charge of 5 and a mass number of 10 . In addition to the alpha particle, a product will be formed of charge \(5-2=3\) and mass number 10 \(-4=6\). This is \({ }_{3}^{6} \mathrm{Li}\). (d) The nuclear charge of the second product particle is \(15-14=\) \(+1\). Its mass number is \(30-30=0\). Hence, the particle must be a positron, \({ }_{+1}^{0} e\). (e) The nuclear charge of the second product particle is \(1-2=-1\). Its mass number is \(3-3=0\). Hence, the particle must be a beta particle (an electron), \({ }_{-1}^{0} e\). (f) The reactants, \({ }_{4}^{9} \mathrm{Be}\) and \({ }_{1}^{1} \mathrm{H}\), have a combined nuclear charge of 22 and mass number of 47 . The ejected product will have charge \(22-21=1\), and mass number \(47-46=1\). This is a proton and should be represented in the parentheses by \(p\). In some of these reactions a neutrino and/or a photon are emitted. We ignore them for this discussion since the charge for both is zero. Moreover, the mass of the photon is zero and the mass of each of the several neutrinos, although not zero, is negligibly small.

Given that Po-210 decays via alpha emission, determine the resulting daughter nucleus.

Uranium-238 \(\left({ }_{92}^{238} \mathrm{U}\right)\) ) is radioactive and decays into a succession of different elements. The following particles are emitted before the nucleus reaches a stable form: \(\alpha, \beta, \beta, \alpha, \alpha, \alpha, \alpha, \alpha, \beta, \beta, \alpha, \beta\), \(\beta\), and \(\alpha\left(\beta\right.\) stands for "beta particle, \(" e^{-}\) ). What is the final stable nucleus? The original nucleus emitted 8 alpha particles and 6 beta particles. When an alpha particle is emitted, \(Z\) decreases by 2 , since the alpha particle carries away a charge of \(+2 e\). A beta particle carries away a charge of \(-1 e\), and so as a result the charge on the nucleus must increase to \((Z+1) e\). We then have, for the final nucleus, Final \(Z=92+6-(2)(8)=82\) Final \(A=238-(6)(0)-(8)(4)=206\) The final stable nucleus is \({ }_{82}^{206} \mathrm{~Pb}\) -

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