Question

Complete and balance the following nuclear equations by supplying the missing particle: (a) \({ }_{58}^{252} \mathrm{Cf}+{ }_{5}^{10} \mathrm{~B} \longrightarrow 3{ }_{0}^{1} \mathrm{n}+\) ? (b) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+\) ? (c) \({ }_{1}^{1} \mathrm{H}+{ }_{5}^{11} \mathrm{~B} \longrightarrow 3\) ? (d) \({ }_{53}^{122} \mathrm{I} \longrightarrow{ }_{54}^{122} \mathrm{Xe}+\) ? (e) \({ }_{26}^{59} \mathrm{Fe} \longrightarrow{ }_{-1}^{0} \mathrm{e}+\) ?

Short answer

(a) \({}_{58}^{252}\mathrm{Cf}+{}_{5}^{10}\mathrm{B} \longrightarrow 3{}_{0}^{1}\mathrm{n}+{}_{63}^{259}\mathrm{Eu}\) (b) \({}_{1}^{2}\mathrm{H}+{}_{2}^{3}\mathrm{He} \longrightarrow{}_{2}^{4}\mathrm{He}+{}_{1}^{1}\mathrm{H}\)

Step-by-step solution

Conservation of Mass Number and Atomic Number

Write down the conservation equations for both mass number (A) and atomic number (Z): A: 252 + 10 = 3(1) + A_missing Z: 58 + 5 = 0 + Z_missing

Solve for the Missing Values

Solve these conservation equations for the missing mass number and atomic number. A: A_missing = 252 + 10 - 3(1) = 259 Z: Z_missing = 58 + 5 = 63

Write the Complete Nuclear Equation

Now that we have the missing mass and atomic numbers, the completed and balanced equation is: \({}_{58}^{252}\mathrm{Cf}+{}_{5}^{10}\mathrm{B} \longrightarrow 3{}_{0}^{1}\mathrm{n}+{}_{63}^{259}\mathrm{?}\) (b) \({}_{1}^{2}\mathrm{H}+{}_{2}^{3}\mathrm{He} \longrightarrow{}_{2}^{4}\mathrm{He}+\) ?

Conservation of Mass Number and Atomic Number

Write down the conservation equations for both mass number (A) and atomic number (Z): A: 2 + 3 = 4 + A_missing Z: 1 + 2 = 2 + Z_missing

Solve for the Missing Values

Solve these conservation equations for the missing mass number and atomic number. A: A_missing = 2 + 3 - 4 = 1 Z: Z_missing = 1 + 2 - 2 = 1

Write the Complete Nuclear Equation

Now that we have the missing mass and atomic numbers, the completed and balanced equation is: \({}_{1}^{2}\mathrm{H}+{}_{2}^{3}\mathrm{He} \longrightarrow{}_{2}^{4}\mathrm{He}+{}_{1}^{1}\mathrm{?}\) Note: The '?' should be replaced with the appropriate chemical symbols, which can be found using the periodic table based on the atomic number. For example, \({}_{63}^{259}\mathrm{?}\) should be replaced by \({}_{63}^{259}\mathrm{Eu}\).

Key concepts

Conservation of Mass Number

In nuclear reactions, one fundamental principle is the conservation of mass number, which states that the sum of mass numbers (the total number of protons and neutrons) in the reactants must equal the sum of the mass numbers in the products. The mass number is denoted as 'A'. To ensure a nuclear equation is balanced regarding mass number, we identify and sum the mass numbers on each side of the reaction and make any necessary adjustments to balance the equation.

For instance, in the example where Californium (Cf) and Boron (B) react to produce neutrons and an unknown element, we use the conservation of mass number to find the mass of the unknown product. The calculation would be as follows: the total mass on the left side is 262 (252 for Cf and 10 for B), and the total mass on the right side is the sum of the mass of the three neutrons (3) plus the mass of the unknown product. Solving for the unknown mass gives us a mass number of 259 for the new element.

Conservation of Atomic Number

Closely related to the conservation of mass number is the conservation of atomic number, which ensures that the number of protons, or the atomic number 'Z', remains constant before and after the reaction. To check for balance regarding atomic number, we follow similar steps as for the mass number, but focus on protons.

In our given example, the atomic number for Californium is 58 and for Boron is 5, giving us a total of 63 on the reactant side. Neutrons do not contribute to the atomic number because they lack a positive charge. Therefore, the atomic number of the unknown element must also be 63. With both mass and atomic numbers in hand, we can now deduce the identity of the unknown product from the periodic table, which, in this case, turns out to be Europium (Eu).

Nuclear Chemistry

Nuclear chemistry is the sub-discipline of chemistry that deals with changes in the nucleus of elements. These changes often involve the transformation of one element into another and the harnessing of the energy changes associated with these transformations. Nuclear chemistry is fundamental in many applications such as energy production in nuclear reactors, medical imaging and treatments, and in understanding cosmic phenomena.

The balancing of nuclear equations is an essential skill within the field of nuclear chemistry because it allows scientists and students to understand the details involved in nuclear reactions, including the types of particles involved, their interactions, and the products formed. This understanding is crucial both for predicting the outcomes of reactions and for ensuring the proper handling of materials, especially when dealing with radioactive substances.

Balancing Nuclear Reactions

Balancing nuclear reactions requires a meticulous approach to ensure that both mass and atomic numbers remain constant from reactants to products. This is crucial for accurately predicting reaction outcomes and adhering to the laws of conservation in nuclear chemistry. To balance nuclear reactions, one must go through a series of steps that involve identifying the starting materials, writing down the conservation equations for both mass number and atomic number, solving for any unknowns, and completing the reaction with appropriate elements or particles while ensuring that the mass and atomic numbers are balanced on both sides of the equation.

The exercises provided demonstrate the application of these principles to solve nuclear equations. The steps guide students to fill in missing particles or unknown elements, reinforcing their understanding of the process. When teaching or learning about balancing nuclear reactions, it is beneficial to refer to a periodic table to identify elements based on their atomic numbers and to practice with a variety of reaction types to gain proficiency in this essential area of nuclear chemistry.