Question

Write balanced equations for (a) \({ }_{92}^{238} \mathrm{U}(\alpha, \mathrm{n}){ }^{241} \mathrm{Pu},\) (b) \({ }^{14} \mathrm{~N}(\alpha, \mathrm{p}){ }^{17} \mathrm{O},(\mathbf{c}){ }_{26}^{56} \mathrm{Fe}\left(\alpha, \beta^{-}\right)_{29}^{60} \mathrm{Cu} .\)

Short answer

The balanced nuclear equations for the given reactions are: (a) \(_{92}^{238} U + _{2}^{4} \alpha \rightarrow_{94}^{241} Pu + _{0}^{1} n\) (b) \(_{7}^{14} N + _{2}^{4} \alpha \rightarrow_{8}^{17} O + _{1}^{1} p\) (c) \(_{26}^{56} Fe + _{2}^{4} \alpha \rightarrow_{28}^{60} Ni + _{0}^{-1} \beta^{-} \rightarrow_{29}^{60} Cu + _{0}^{-1} \beta^{-}\)

Step-by-step solution

Write the equation for \( _{92}^{238} U (\alpha, n) { }_{94}^{241} Pu\)

We start with the Uranium-238 nucleus and the alpha particle: \(_{92}^{238} U + _{2}^{4} \alpha\) After the alpha particle is absorbed, a neutron is emitted: \(_{92}^{238} U + _{2}^{4} \alpha \rightarrow_{94}^{241} Pu + _{0}^{1} n\) Hence, the balanced equation for the reaction is: \(_{92}^{238} U + _{2}^{4} \alpha \rightarrow_{94}^{241} Pu + _{0}^{1} n\) (b)

Write the equation for \( _{7}^{14} N (\alpha, p) { }_{8}^{17} O\)

We start with the Nitrogen-14 nucleus and the alpha particle: \(_{7}^{14} N + _{2}^{4} \alpha\) After the alpha particle is absorbed, a proton is emitted: \(_{7}^{14} N + _{2}^{4} \alpha \rightarrow_{8}^{17} O + _{1}^{1} p\) Hence, the balanced equation for the reaction is: \(_{7}^{14} N + _{2}^{4} \alpha \rightarrow_{8}^{17} O + _{1}^{1} p\) (c)

Write the equation for \(_{26}^{56} Fe (\alpha, \beta^{-}){ }_{29}^{60} Cu\)

We start with the Iron-56 nucleus and the alpha particle: \(_{26}^{56} Fe + _{2}^{4} \alpha\) After the alpha particle is absorbed, a beta-minus particle is emitted: \(_{26}^{56} Fe + _{2}^{4} \alpha \rightarrow_{28}^{60} Ni + _{0}^{-1} \beta^{-}\) Finally, the beta-minus decay takes place, converting a neutron in the Nickel-60 nucleus into a proton and emitting an electron, forming Copper-60: \(_{28}^{60} Ni \rightarrow_{29}^{60} Cu + _{0}^{-1} \beta^{-}\) So, combining these reactions we get the overall equation for the reaction: \(_{26}^{56} Fe + _{2}^{4} \alpha \rightarrow_{28}^{60} Ni + _{0}^{-1} \beta^{-} \rightarrow_{29}^{60} Cu + _{0}^{-1} \beta^{-}\) Balancing these equations ensures that mass and atomic numbers are conserved throughout each nuclear reaction.

Key concepts

Nuclear Equations

Nuclear equations are similar to chemical equations but pertain to changes within an atom's nucleus. In such equations, it is essential to balance both the atomic and mass numbers. This guarantees the law of conservation of mass and energy is upheld.
Here's what you need to consider:

• The atomic number indicates the number of protons in the nucleus. It appears as a subscript before the element symbol.
• The mass number shows the total of protons and neutrons, depicted as a superscript.
• During a nuclear reaction, an atom may benefit from an increase, decrease, or transformation into another isotope or element.

Consider a simplified equation:
\(^{A}_{Z}X + ^{4}_{2}\alpha \rightarrow ^{A}_{Z}Y + ^{A'}_{Z'}N\). Here, \(X\) is the initial element, \(\alpha\) is the alpha particle, \(Y\) is the product after decay or transformation, and \(N\) represents emissions, like neutrons.

Alpha Decay

Alpha decay refers to a process where an unstable nucleus ejects an alpha particle. An alpha particle is composed of two protons and two neutrons, equivalent to a helium nucleus.
This decay reduces the original nucleus's atomic number by 2 and its mass number by 4. It's a common way for heavy elements to reduce excess mass and become more stable.
Here’s a quick breakdown:

• Start with your original heavy nucleus, such as Uranium-238.
• After emitting an alpha particle, you get a nucleus with reduced atomic and mass numbers, like Thorium-234.

In equation form:\(^{238}_{92}\mathrm{U} \rightarrow ^{234}_{90}\mathrm{Th} + ^{4}_{2}\alpha\).
Alpha decay ensures mass reduction, helping heavy isotopes stabilize.

Beta Decay

Beta decay is a fascinating nuclear reaction that involves the transformation of a neutron into a proton with the emission of a beta particle (an electron or positron). There are two types, beta-minus (\(\beta^{-}\)) and beta-plus (\(\beta^{+}\)). We're focusing on \(\beta^{-}\) here:

• During beta-minus decay, a neutron is converted to a proton.
• This results in an electron (\(\beta^{-}\)) being emitted along with an antineutrino.
• Atomic number increases by 1 as a new element is formed.

For example, let's look at Nickel-60 undergoing beta-minus decay:\(^{60}_{28}\mathrm{Ni} \rightarrow ^{60}_{29}\mathrm{Cu} + ^{0}_{-1}\beta^{-}\).
In a nuclear reaction equation, this process helps convert heavier isotopes to their more stable forms, crucial for elements' transformation into stable forms.

Isotopes

Isotopes are variants of a particular element that share the same number of protons but have different numbers of neutrons, giving them different mass numbers.
Isotopes are incredibly useful both in natural processes and various applications like medicine and archaeology. Here's what they entail:

• Each isotope of an element has identical chemical properties but differing nuclear behaviors.
• They are essential in nuclear reactions, where nuclei may transform into other isotopes or entirely different elements.
• The mass number, shown as a superscript, varies among isotopes, but the elemental identity (atomic number) remains the same.

For instance, Carbon-12, Carbon-13, and Carbon-14 are isotopes of carbon. They are vital tools, such as Carbon-14 used in radiocarbon dating. Understanding isotopes is key to mastering nuclear reactions and identifying balanced nuclear equations.