Question

Balance the following nuclear reactions: a. \(\quad_{93}^{239} \mathrm{Np} \longrightarrow-1^{0} \beta+\underline{?} b. \)\quad_{4}^{9} \mathrm{Be}+_{2}^{4} \mathrm{He} \longrightarrow \underline{?}\( c. \)\quad \frac{32}{15} \mathrm{P}+\underline{?} \longrightarrow_{15}^{33} \mathrm{P}\( d. \)\quad_{92}^{236} \mathrm{U} \longrightarrow_{36}^{94} \mathrm{Kr}+\underline{?}+3_{0}^{1} n$

Short answer

a. \({}_{93}^{239} \mathrm{Np}\rightarrow_{-1}^{0} \beta +_{94}^{239} \mathrm{Pu}\)\ b. \({}_{4}^{9} \mathrm{Be}+_{2}^{4} \mathrm{He}\rightarrow_{6}^{13} \mathrm{C}\)\ c. \({}_{15}^{32} \mathrm{P}+_{0}^{1} \mathrm{n}\rightarrow _{15}^{33} \mathrm{P}\) d. \({}_{92}^{236} \mathrm{U}\rightarrow_{36}^{94} \mathrm{Kr}+_{56}^{139} \mathrm{Ba}+3_{0}^{1} \mathrm{n}\)

Step-by-step solution

- Balance Reaction (a)

In this reaction, \({}_{93}^{239} \mathrm{Np}\rightarrow_{-1}^{0} \beta + \underline{?}\) \(\beta\) particle (beta particle) has an atomic number of -1 and a mass number of 0. To find the missing element, balance both atomic number and mass number: 239 (Np mass number) = 0 (beta mass number) + Mass number of \ \underline{?}. So, the mass number of the unknown element remains 239. Next, balance the atomic numbers: 93 (Np atomic number) = -1 (beta atomic number) + Atomic number of \ \underline{?}. So, the new atomic number is 94. The element with atomic number 94 is Plutonium (Pu). Thus, the reaction becomes \ \({}_{93}^{239} \mathrm{Np}\rightarrow_{-1}^{0} \beta +_{94}^{239} \mathrm{Pu}\).

- Balance Reaction (b)

This reaction is given by \({}_{4}^{9} \mathrm{Be}+_{2}^{4} \mathrm{He}\rightarrow_\underline{?}\). First, sum the mass numbers on the left side: 9 (Be) + 4 (He) = 13. So, the mass number of the product is 13. Next, sum the atomic numbers: 4 (Be) + 2 (He) = 6. Therefore, the missing element has an atomic number of 6. The element with atomic number 6 is Carbon (C). The balanced reaction is \({}_{4}^{9} \mathrm{Be}+_{2}^{4} \mathrm{He}\rightarrow_{6}^{13} \mathrm{C}\).

- Balance Reaction (c)

The given reaction is \({}_{15}^{32} \mathrm{P}+\underline{?} \rightarrow _{15}^{33} \mathrm{P}\). To balance the mass numbers: 32 (P) + Mass number of \underline{?} = 33. Thus, the mass number of the unknown particle is 1. Next, balance the atomic numbers: 15 (P) + Atomic number of \underline{?} = 15. Thus, the atomic number of the unknown particle is 0. The particle that fits this description is a neutron \(_{0}^{1} n\). The balanced reaction is \({}_{15}^{32} \mathrm{P}+_{0}^{1} \mathrm{n}\rightarrow _{15}^{33} \mathrm{P}\).

- Balance Reaction (d)

The reaction provided is \({}_{92}^{236} \mathrm{U}\rightarrow_{36}^{94} \mathrm{Kr}+\underline{?}+3_{0}^{1} \mathrm{n}\). First, balance the mass numbers: 236 (U) = 94 (Kr) + Mass number of \underline{?} + 3 (neutrons). Therefore, the mass number of the unknown element is 236 - 94 - 3 = 139. Next, balance atomic numbers: 92 (U) = 36 (Kr) + Atomic number of \underline{?} + 0 (neutrons). Thus, the atomic number of the unknown element is 92 - 36 = 56. The element with atomic number 56 is Barium (Ba). Therefore, the balanced reaction is \({}_{92}^{236} \mathrm{U}\rightarrow_{36}^{94} \mathrm{Kr}+_{56}^{139} \mathrm{Ba}+3_{0}^{1} \mathrm{n}\).

Key concepts

Mass Number

The mass number of an atom is the total number of protons and neutrons in its nucleus. It's denoted as the superscript on the left of an element's symbol, for example, \(_{4}^{9} \text{Be}\) where 9 is the mass number.
When balancing nuclear reactions, the sum of the mass numbers (protons + neutrons) on the reactant side must equal the sum on the product side. This conservation principle helps in determining unknown particles or elements in nuclear reactions.
For instance, in the reaction \(\text{}_{93}^{239} \text{Np} \rightarrow _{-1}^{0} \beta + \text{}_{94}^{239} \text{Pu}\), the mass number remains balanced as 239 on both sides.

Atomic Number

The atomic number is the number of protons in an atom's nucleus and gives an element its identity. It is noted as the subscript before an element's symbol, such as \(_{6} \text{C}\) where 6 is the atomic number of carbon.
While balancing nuclear reactions, the sum of atomic numbers on the reactant side should equal those on the product side to ensure charge conservation. For example, in \(\text{}_{92}^{236} \text{U} \rightarrow _{36}^{94} \text{Kr} + \text{}_{56}^{139} \text{Ba} + 3 _{0}^{1} n\), the sum of atomic numbers 92 (U) is equal to 36 (Kr) + 56 (Ba).
This rule helps to identify unknown elements in a reaction and ensure the reaction adheres to the conservation of atomic charge.

Beta Particle

A beta particle is represented by \text{\(_{-1}^{0} \beta\)} and is essentially a high-energy, high-speed electron or positron emitted during radioactive decay.
It has negligible mass (\text{mass number} = 0) and an electric charge of \text{-1} or \text{+1} depending on whether it is an electron or positron. Beta decay is a process that results in the transformation of a neutron into a proton, emitting a beta particle and an antineutrino (\text{for \(\beta^{-}\) decay}) or the transformation of a proton into a neutron, emitting a positron and a neutrino (\text{for \(\beta^{+}\) decay}).
For example, in \text{\(_{93}^{239} \text{Np} \rightarrow _{-1}^{0} \beta + _{94}^{239} \text{Pu}\)}, a neutron in \text{\(_{93}^{239} \text{Np}\)} decays into a proton forming \text{\(_{94}^{239} \text{Pu}\)} and emitting a beta particle.

Element Identification

Element identification in nuclear reactions involves using the atomic number to determine which element is present. Each element on the periodic table has a unique atomic number which helps in identifying it.
For example, the element with the atomic number 6 is Carbon (C) and the one with the atomic number 94 is Plutonium (Pu).
In a nuclear reaction such as \text{\(_{15}^{32} \text{P} + _{0}^{1} \text{n} \rightarrow _{15}^{33} \text{P}\)}, the atomic number remains 15, indicating the element is Phosphorus (P) despite the neutron addition. Similarly, to identify an element produced in reactions, knowing the atomic number helps us trace back to its position on the periodic table.
This principle ensures that even in complex reactions, we can correctly identify and balance the various elements involved.