Question

Complete and balance the following nuclear equations by supplying the missing particle: \begin{equation}\begin{array}{l}{\text { (a) }_{98}^{252} \mathrm{Cf}+_{5}^{10} \mathrm{B} \longrightarrow 3_{0}^{1} \mathrm{n}+?} \\\ {\text { (b) }_{1}^{2} \mathrm{H}+_{2}^{3} \mathrm{He} \longrightarrow_{2}^{4} \mathrm{He}+?}\\\ {\text { (c) }_{1}^{1} \mathrm{H}+_{5}^{11} \mathrm{B} \longrightarrow 3?} \\ {\text { (d) }_{53}^{122} \mathrm{I}\longrightarrow_{54}^{122} \mathrm{Xe}+?}\\\ {\text { (e) }_{26}^{59} \mathrm{Fe}\longrightarrow_{-1}^{0} \mathrm{e}+?}\end{array}\end{equation}

Short answer

Here are the short answers to the given nuclear reactions: (a) \(\ce{_{98}^{252}Cf + _5^{10}B -> 3_0^1n + _{103}^{259}Lr}\) (b) \(\ce{_{1}^{2}H + _2^{3}He -> _2^{4}He + _1^1H}\) (c) \(\ce{_{1}^{1}H + _5^{11}B -> 3_{2}^{4}He}\) (d) \(\ce{_{53}^{122}I -> _{54}^{122}Xe + _{-1}^0e}\) (e) \(\ce{_{26}^{59}Fe -> _{-1}^{0}e + _{27}^{59}Co}\)

Step-by-step solution

(a) Reaction: Cf + B -> 3n + ?

First, let's write down the given reaction: \[\ce{_{98}^{252}Cf + _5^{10}B -> 3_0^1n + ?}\] To find the missing particle, we need to ensure the conservation of mass number and atomic number: Total mass number on left side = 252 + 10 = 262 Total atomic number on left side = 98 + 5 = 103 Now, there are 3 neutrons on the right side: Mass number accounted for on right side = 3 * 1 = 3 Atomic number accounted for on right side = 0 (Since neutrons have an atomic number of 0) For conservation of mass number: 262 - 3 = 259 For conservation of atomic number: 103 - 0 = 103 So, the missing particle is \(\ce{_{103}^{259}Lr}\). The balanced equation is: \[\ce{_{98}^{252}Cf + _5^{10}B -> 3_0^1n + _{103}^{259}Lr}\]

(b) Reaction: H + He -> He + ?

Write down the given reaction: \[\ce{_{1}^{2}H + _2^{3}He -> _2^{4}He + ?}\] To find the missing particle, we need to ensure the conservation of mass number and atomic number: Total mass number on left side = 2 + 3 = 5 Total atomic number on left side = 1 + 2 = 3 On the right side we have: Mass number accounted for = 4 Atomic number accounted for = 2 For conservation of mass number: 5 - 4 = 1 For conservation of atomic number: 3 - 2 = 1 So, the missing particle is \(\ce{_{1}^{1}H}\) (a proton). The balanced equation is: \[\ce{_{1}^{2}H + _2^{3}He -> _2^{4}He + _1^1H}\]

(c) Reaction: H + B -> 3?

Write down the given reaction: \[\ce{_{1}^{1}H + _5^{11}B -> 3?}\] Total mass number on left side = 1 + 11 = 12 Total atomic number on left side = 1 + 5 = 6 Since there are 3 identical particles on the right side, let's denote the mass number of each as m and atomic number as a: For conservation of mass number: 12 = 3 * m For conservation of atomic number: 6 = 3 * a Solving these equations, we get m = 4 and a = 2. Thus, the missing particle is \(\ce{_{2}^{4}He}\). The balanced equation is: \[\ce{_{1}^{1}H + _5^{11}B -> 3_{2}^{4}He}\]

(d) Reaction: I -> Xe + ?

Write down the given reaction: \[\ce{_{53}^{122}I -> _{54}^{122}Xe + ?}\] Total mass number on left side = 122 Total atomic number on left side = 53 On the right side, we have: Mass number accounted for = 122 Atomic number accounted for = 54 For conservation of mass number: 122 - 122 = 0 For conservation of atomic number: 53 - 54 = -1 So, the missing particle is \(\ce{_{-1}^{0}e}\) (an electron). The balanced equation is: \[\ce{_{53}^{122}I -> _{54}^{122}Xe + _{-1}^0e}\]

(e) Reaction: Fe -> e + ?

Write down the given reaction: \[\ce{_{26}^{59}Fe -> _{-1}^{0}e + ?}\] Total mass number on left side = 59 Total atomic number on left side = 26 On the right side, we have: Mass number accounted for = 0 Atomic number accounted for = -1 For conservation of mass number: 59 - 0 = 59 For conservation of atomic number: 26 - (-1) = 27 So, the missing particle is \(\ce{_{27}^{59}Co}\). The balanced equation is: \[\ce{_{26}^{59}Fe -> _{-1}^{0}e + _{27}^{59}Co}\]

Key concepts

Mass Number Conservation

Nuclear reactions are fascinating processes that see the transformation of atomic nuclei. In these reactions, the mass number must remain constant. This principle is known as mass number conservation. The mass number of an atom is the sum of its protons and neutrons. During a nuclear reaction, the total mass number of the reactants before the reaction must equal the total mass number of the products.

For instance, in the given exercise, you can see this principle at work when calculating the reactions. Let's take Step 1, the reaction of Californium (Cf) with Boron (B): the mass number on the left side totals 262, while on the right side, after three neutrons are accounted for, the remaining mass number is 259. Hence, the missing particle must have a mass number of 259 to balance the equation, ensuring mass number conservation.

Atomic Number Conservation

Another crucial principle in nuclear reactions is atomic number conservation. The atomic number represents the number of protons in an atom's nucleus and defines the element. During a nuclear reaction, the sum of atomic numbers of the reactants must equal the sum of atomic numbers of the products.

Referring to the original solution, in Step 2, the reaction involving Deuterium (H) and Helium (He), we assess atomic numbers: 1 from H and 2 from He gives a total of 3 on the reactant side. For the products, Helium contributes an atomic number of 2, leaving a requirement of an additional particle with an atomic number of 1. Hence, the additional particle is a proton (\(_{1}^{1}H\)), balancing the atomic numbers on both sides.

Nuclear Equations

Nuclear equations are much like chemical equations but focus on changes in the nucleus. Balancing nuclear equations involves ensuring mass number and atomic number conservation. This balance helps identify missing particles or isotopes in a nuclear transformation.

In the exercise, nuclear equations are given where certain nuclear transformations are happening. For instance, in Step 3 of the given solution, combining Hydrogen (H) and Boron (B), results in three Helium particles. The equation \(_{1}^{1}H + _5^{11}B \rightarrow 3_{2}^{4}He\) is perfectly balanced, maintaining the sum of mass numbers and atomic numbers on both sides. This guarantees that mass is neither lost nor created, satisfying the principles central to nuclear equations.

Nuclear Particles

In nuclear reactions, several types of particles are involved, including protons, neutrons, and occasionally electrons or other exotic particles such as alpha or beta particles. Understanding these particles helps in determining the outcome of nuclear reactions.

From the provided exercise, it is evident that identifying these nuclear particles is critical. For example, in Step 4, where Iodine (I) decays into Xenon (Xe), the difference in atomic numbers is -1, indicating the involvement of a beta particle (electron), \(_{-1}^{0}e\). Recognizing these particles allows us to deduce unknown components in nuclear reactions and ensures the reaction obeys nuclear conservation laws.