Problem 51 - Chemistry and Chemical Reactivity

Question

Some of the reactions explored by Rutherford and others are listed below. Identify the unknown species in each reaction. (a) \(^{14}_{7} \mathrm{p}+_{2}^{4} \mathrm{He} \longrightarrow_{8}^{17} \mathrm{O}+?\) (b) \(_{4}^{9} \mathrm{Be}+_{2}^{4} \mathrm{He} \longrightarrow ?+_{0}^{1} \mathrm{n}\) (c) \(?+_{2}^{4} \mathrm{He} \longrightarrow_{15}^{30} \mathrm{P}+_{0}^{1} \mathrm{n}\) (d) \(^{239}_{94} \mathrm{Pu}+_{2}^{4} \mathrm{He} \longrightarrow ?+_{0}^{1} \mathrm{n}\)

Step-by-Step Solution

Answer

(a) \( ^{1}_{1}\mathrm{p} \), (b) \( _{6}^{12}\mathrm{C} \), (c) \( _{13}^{27}\mathrm{Al} \), (d) \( _{96}^{242}\mathrm{Cm} \).

Step 1: Reaction (a) - Subtract to find unknown

In the reaction \( ^{14}_{7} \mathrm{N} + _{2}^{4} \mathrm{He} \longrightarrow _{8}^{17} \mathrm{O} + ? \), apply the conservation of mass number and atomic number. The mass number before and after the reaction must be equal as well as the atomic number.**Mass numbers:** \(14 + 4 = 17 + A?\) leading to \(A? = 1\).**Atomic numbers:** \(7 + 2 = 8 + Z?\) leading to \(Z? = 1\).This implies the unknown is a proton (\( ^{1}_{1}\mathrm{p} \)).

Step 2: Reaction (b) - Apply conservation laws

For the reaction \( _{4}^{9} \mathrm{Be} + _{2}^{4} \mathrm{He} \longrightarrow ? + _{0}^{1} \mathrm{n} \), use the conservation laws.**Mass numbers:** \(9 + 4 = A? + 1\) leading to \(A? = 12\).**Atomic numbers:** \(4 + 2 = Z? + 0\) leading to \(Z? = 6\).Thus, the unknown is \( _{6}^{12}\mathrm{C} \) (Carbon).

Step 3: Reaction (c) - Reverse solving for unknown

Given \( ? + _{2}^{4} \mathrm{He} \longrightarrow _{15}^{30} \mathrm{P} + _{0}^{1} \mathrm{n} \), we find the reactant by reversing the conservation equations.**Mass numbers:** \(A? + 4 = 30 + 1\) leading to \(A? = 27\).**Atomic numbers:** \(Z? + 2 = 15 + 0\) leading to \(Z? = 13\).Therefore, the unknown species is \( _{13}^{27}\mathrm{Al} \) (Aluminium).

Step 4: Reaction (d) - Identify unknown species

For \( ^{239}_{94} \mathrm{Pu} + _{2}^{4} \mathrm{He} \longrightarrow ? + _{0}^{1} \mathrm{n} \), follow the conservation.**Mass numbers:** \(239 + 4 = A? + 1\) resulting in \(A? = 242\).**Atomic numbers:** \(94 + 2 = Z? + 0\) leading to \(Z? = 96\).Thus, the unknown is \( _{96}^{242}\mathrm{Cm} \) (Curium).

Key Concepts

Conservation of Mass Number

The conservation of mass number is an essential concept in nuclear reactions. This principle states that the total mass number (the sum of protons and neutrons) must remain the same before and after a reaction. This allows us to balance nuclear equations effectively.
For instance, in the case of reaction (a), where nitrogen-14 is involved with helium-4 to produce oxygen-17, we used this law to find that the missing particle is a proton. We calculated the mass numbers as follows:

• The sum of the mass numbers on the left side of the equation: 14 (nitrogen) + 4 (helium) = 18.
• The sum on the right side: 17 (oxygen) + X = 18.

Here, X must be 1, indicating the presence of a proton. This example illustrates the significance of conserving the mass number to identify unknown species in nuclear reactions.

Conservation of Atomic Number

The conservation of atomic number is another vital aspect of nuclear reactions. This rule ensures the total number of protons, represented by the atomic number, remains constant through the reaction. This aids us in predicting and balancing nuclear changes.
In reaction (a), after applying the conservation of atomic number:

• Adding the atomic numbers on the reactants side: 7 (nitrogen) + 2 (helium) = 9.
• Which must equal the atomic numbers on the products side: 8 (oxygen) + Z.
• This reveals that Z is 1.

Thus, we identified the emitted particle as a proton (symbolized as \(^{1}_{1} ext{p} \). This principle confirms that nuclear reactions rearrange but do not create or destroy protons.

Proton Emission

Proton emission occurs when a nucleus releases a proton, resulting in a decrease in its atomic number by one unit while the mass number remains unchanged. This is less common but important in identifying elements and balancing nuclear reactions.
Take reaction (a) as an example again; the identification of the unknown as a proton signifies proton emission. This results in the transformed nucleus (oxygen) having one less proton compared to the combined nuclei before the reaction.

• This adjustment in atomic structure is crucial for altering the identity of the involved element.
• Therefore, understanding proton emission helps us predict outcomes and confirm the products formed in nuclear reactions.

Neutron Emission

Neutron emission is a type of nuclear reaction where a neutron is released from the nucleus. This event doesn’t change the atomic number but reduces the mass number by one.
In step (b), after completing the necessary conservation calculations, we observed this in action. When beryllium-9 and helium-4 react, they produce carbon-12 and emit a neutron.

• Thus reflecting a decrease in mass number while retaining the same atomic number for the newly formed element.
• Similarly, steps related to reactions (c) and (d) also show neutron emissions, helping in balancing the nuclear equations and identifying the resultant elements.

Neutron emissions are instrumental in stabilizing the nucleus as well as detailing transformations in nuclear reactions.