Complete the following nuclear reactions : (a) \({ }_{42}^{~} \mathrm{Mo}(\ldots, n){ }_{43}^{97} \mathrm{Tc}\) (b) .... \((\alpha, 2 n) \stackrel{211}{85} \mathrm{At}\) (c) \({ }_{25}^{55} \mathrm{Mn}(n, \gamma) \ldots\) (d) \({ }_{96}^{246} \mathrm{Cm}+{ }_{6}^{12} \mathrm{C} \longrightarrow \ldots .+4_{0}^{1} n\) (e) \({ }_{13}^{27} \mathrm{Al}(\alpha, n) \ldots\) (f) \({ }_{92}^{215} \mathrm{U}\left(\alpha, \beta^{-}\right) \ldots\)

Short Answer

Expert verified
The complete nuclear reactions are: (a) \({ }_{42}^{96} \text{Mo}(p, n){ }_{43}^{97} \text{Tc}\), (b) \({ }_{83}^{209} \text{Bi}(\text{α}, 2 n){ }_{85}^{211} \text{At}\), (c) \({ }_{25}^{55} \text{Mn}(n, \text{γ}){ }_{25}^{56} \text{Mn}\), (d) \({ }_{96}^{246} \text{Cm}+ { }_{6}^{12} \text{C} \longrightarrow { }_{102}^{254} \text{No} + 4_{0}^{1} n\), (e) \({ }_{13}^{27} \text{Al}(\text{α}, n){ }_{14}^{30} \text{Si}\), and (f) \({ }_{92}^{215} \text{U}(\text{α}, \text{β}^{-}){ }_{95}^{219} \text{Am}\).

Step by step solution

01

Understanding Nuclear Reactions

A nuclear reaction involves a change in an atom's nucleus and generally results in the change of one element into another. The reaction is typically written with an incoming particle and an ejected particle, alongside the starting and resulting isotopes. During a nuclear reaction, both the atomic number (proton count, denoted as Z) and mass number (nucleon count, denoted as A) are conserved. For emissions, the mass and atomic numbers of the product are reduced by the mass and atomic numbers of the emitted particle. For absorptions, the mass and atomic numbers of the product increase by the mass and atomic numbers of the absorbed particle.
02

Determine the incoming particle in Reaction (a)

For the reaction \[\begin{equation}_{42}^{~} \text{Mo}(\text{...}, n)_{43}^{97} \text{Tc}, \end{equation}\] we know a neutron ('n') is the ejected particle. To conserve the nucleon number, the mass number must remain the same, as neutron emission does not change it. Since the atomic number increases from 42 to 43, the incoming particle must have a +1 atomic number to raise the atomic number of Molybdenum (Mo) to that of Technetium (Tc), which means the incoming particle is a proton (p). Thus, the complete reaction is \[\begin{equation}_{42}^{96} \text{Mo}(p, n)_{43}^{97} \text{Tc}.\end{equation}\]
03

Identify the starting isotope in Reaction (b)

For the reaction \[\begin{equation}\ldots (\text{α}, 2 n)_{85}^{211} \text{At},\end{equation}\] we know an alpha particle ('α') is absorbed and two neutrons ('2n') are ejected. An alpha particle has an atomic number of 2 and a mass number of 4. The resulting isotope has an atomic number of 85 and a mass number of 211. Therefore, the starting isotope must have an atomic number of 85 - 2 = 83 (bismuth, Bi) and a mass number of 211 - 4 + 2 = 209. Hence, the complete reaction is \[\begin{equation}_{83}^{209} \text{Bi}(\text{α}, 2 n)_{85}^{211} \text{At}.\end{equation}\]
04

Calculate the resulting isotope in Reaction (c)

For the reaction \[\begin{equation}_{25}^{55} \text{Mn}(n, \text{γ}) \ldots,\end{equation}\] a neutron is absorbed and a gamma ray ('γ') is emitted. Gamma emission does not change the atomic or mass numbers. Since a neutron has a mass number of 1, the result will have a mass number increased by 1 to 55 + 1 = 56, with the atomic number remaining the same at 25. Thus, the resulting isotope is \[\begin{equation}_{25}^{56} \text{Mn}.\end{equation}\]
05

Deduce the resulting element in Reaction (d)

The reaction is a fusion between \[\begin{equation}_{96}^{246} \text{Cm} + _{6}^{12} \text{C} \longrightarrow \ldots .+ 4_{0}^{1} n.\end{equation}\] Fusion combines nuclei, and we eject four neutrons. Summing the atomic numbers, we have 96 + 6 = 102, and summing the mass numbers while subtracting four neutrons, we have 246 + 12 - 4 = 254. The resulting element with atomic number 102 is Nobelium (No), and the completed reaction is \[\begin{equation}_{96}^{246} \text{Cm} + _{6}^{12} \text{C} \longrightarrow _{102}^{254} \text{No} + 4_{0}^{1} n.\end{equation}\]
06

Complete the transmutation in Reaction (e)

In the reaction \[\begin{equation}_{13}^{27} \text{Al}(\text{α}, n) \ldots,\end{equation}\] we absorb an alpha particle and emit a neutron. An alpha particle has atomic and mass numbers of 2 and 4 respectively, while a neutron has mass number 1. The resulting element has an atomic number of 13 + 2 - 1 = 14 (which is Silicon, Si) and a mass number of 27 + 4 - 1 = 30. The complete reaction is \[\begin{equation}_{13}^{27} \text{Al}(\text{α}, n)_{14}^{30} \text{Si}.\end{equation}\]
07

Finalize the sequence in Reaction (f)

For \[\begin{equation}_{92}^{215} \text{U}(\text{α}, \text{β}^{-}) \ldots,\end{equation}\] we're absorbing an alpha particle and emitting a beta minus particle (an electron), which means a neutron is converting into a proton within the nucleus. Absorption of the alpha particle (2 protons, 4 nucleons) increases the atomic number by 2 and mass number by 4. Emission of a beta particle increases the atomic number by 1 (as one more proton is formed), but does not change the mass number. So, the final element will have an atomic number of 92 + 2 + 1 = 95 (which is Americium, Am) and a mass number of 215 + 4 = 219. The complete reaction is \[\begin{equation}_{92}^{215} \text{U}(\text{α}, \text{β}^{-})_{95}^{219} \text{Am}.\end{equation}\]

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Key Concepts

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

Nuclear Transmutation
Nuclear transmutation is the process by which the nucleus of an atom is changed, resulting in the conversion of one element into another. This remarkable process can occur naturally, such as in radioactive decay, or can be induced artificially in laboratories and nuclear reactors. For students, the concept might appear daunting at first, but understanding it is crucial for grasping the fundamental principles of nuclear chemistry.

For instance, when a neutron is absorbed by an atom and a different particle is ejected, such as in some of the reactions provided in the exercise, the nucleus of the atom undergoes a transformation. This can cause changes in the proton number, which by definition means we've transitioned into a different element. In everyday language, it's somewhat like changing the identity of the atom. This is why transmutation is key for various applications, including the synthesis of new elements and isotopes with purposes ranging from medicine to energy production.

Nobody should let the complexity hide the simplicity of the concept's core: transmutation is just about atoms changing their 'clothes' and becoming something new, triggered by particles that come knocking on their 'door'.
Isotopic Notation
Understanding isotopic notation is akin to learning the language of nuclear chemistry. When we write an isotope, we use a specific format: the element's symbol, preceded by its atomic mass number (A) as a superscript and its atomic number (Z) as a subscript. For example, carbon-12 is written as \( _{6}^{12}C \).

Here are the basics to remember: the atomic number (Z) indicates the number of protons in the nucleus, which defines the element itself, and the mass number (A) is the sum of protons and neutrons, which tells us the isotope of that element. Why is this important? Well, isotopes of an element have the same number of protons but a different number of neutrons, which means they can behave very differently in terms of stability and reactivity.

  • Atomic number (subscript) - Identity Card of the atom
  • Mass number (superscript) - tells us 'how heavy' the atom is with all its protons and neutrons
Now, why not look at the isotopic notation as a shorthand or a tweet that conveys critical information about an atom's identity and weight? Just remember the position of the numbers in the notation, and you're on your way to mastering the language of atoms!
Conservation Laws in Nuclear Chemistry
The conservation laws in nuclear chemistry are the fundamental 'rules of the game' that dictate how nuclear reactions can and cannot happen. Simply put, these laws tell us that during any nuclear process, certain properties must remain constant. There are two main conservation laws to keep in mind: the conservation of mass number and the conservation of charge.

The law of conservation of mass number states that the total number of nucleons (protons + neutrons) must remain the same before and after the reaction. It's like making sure that every person who entered a party also leaves it - nobody disappears into thin air! Similarly, the law of conservation of charge ensures that the total charge remains unaffected by a nuclear reaction. A party where every guest keeps hold of their 'plus one', so to speak.

Conserving More Than Just Numbers

It's not just about keeping score of protons and neutrons. These laws underpin everything in nuclear chemistry and are essential for the stability of atoms and the entire universe. By following these laws, physicists and chemists can predict the products of nuclear reactions and understand the mechanisms behind energy release in processes like nuclear fission and fusion.

  • Conservation of mass number - Keeping track of every neutron and proton
  • Conservation of charge - Balancing the nucleus's charge in the atomic dance
Once you get these conservation laws down, you're well equipped to handle any nuclear equation that comes your way!

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