Question

SOLUTION Analyze We are asked to determine the nucleus that results when radium-226 loses an alpha particle. Plan We can best do this by writing a balanced nuclear reaction for the process. Solve The periodic table shows that radium has an atomic number of 88 . The complete chemical symbol for radium- 226 is therefore \({ }_{85}^{226} \mathrm{Ra}\). An alpha particle is a helium-4 nucleus, and so its symbol is \({ }_{2}^{4} \mathrm{He}\). The alpha particle is a product of the nuclear reaction, and so the equation is of the form $$ { }_{8}^{226} \mathrm{Ra} \longrightarrow{ }_{2}^{A} \mathrm{X}+{ }_{2}^{4} \mathrm{He} $$ where \(A\) is the mass number of the product nucleus and \(Z\) is its atomic number. Mass numbers and atomic numbers must balance, so $$ 226=A+4 $$ and $$ 88=Z+2 $$ Hence, $$ A=222 \text { and } Z=86 $$ Again, from the periodic table, the element with \(Z=86\) is radon (Rn). The product, therefore, is \({ }_{86}^{222} \mathrm{Rn}\), and the nuclear equation is $$ { }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} $$

Short answer

The resulting nucleus after radium-226 loses an alpha particle is radon-222, which is represented by the balanced nuclear reaction equation: \( _{88}^{226}Ra \rightarrow _{86}^{222}Rn + _{2}^{4}He \).

Step-by-step solution

Identify the initial nucleus

We know that the initial nucleus is radium-226, which has an atomic number of 88. Thus, it is represented as \( _{88}^{226}Ra \).

Identify what it loses, the alpha particle

An alpha particle is a helium-4 nucleus, having mass number 4 and atomic number 2. It is represented as \( _{2}^{4}He \).

Write the nuclear reaction equation

Let the resulting nucleus be \( _{Z}^{A}X \), where X represents the new element. Now, according to the given information, our nuclear reaction equation is: \[ _{88}^{226}Ra \rightarrow _{Z}^{A}X + _{2}^{4}He \]

Apply conservation of mass numbers and atomic numbers

To balance the nuclear reaction equation, we must have the same mass numbers and atomic numbers on both sides of the equation: 1. Conservation of mass numbers: \[ 226 = A + 4 \] 2. Conservation of atomic numbers: \[ 88 = Z + 2 \]

Solve for A and Z

From the conservation equations, we can find the values of A and Z: 1. Solving for A: \[ A = 226 - 4 = 222 \] 2. Solving for Z: \[ Z = 88 - 2 = 86 \]

Identify the resulting element and write the balanced equation

Now that we have the atomic number of the resulting nucleus (Z = 86), we can use the periodic table to identify that the element is radon (Rn). Thus, the balanced nuclear reaction equation becomes: \[ _{88}^{226}Ra \rightarrow _{86}^{222}Rn + _{2}^{4}He \] The resulting nucleus after radium-226 loses an alpha particle is radon-222.

Key concepts

Alpha Decay

Alpha decay is a type of nuclear reaction, where an unstable nucleus releases a particle comprised of two protons and two neutrons. This particle is known as an alpha particle.
An alpha particle can be symbolized as \(^4_2 He\), representing a helium-4 nucleus.
During alpha decay, the original nucleus loses these protons and neutrons, causing a decrease in both its atomic and mass numbers.

• The atomic number decreases by 2, suggesting the loss of two protons.
• The mass number decreases by 4, corresponding to the loss of two protons and two neutrons.

These changes result in the transformation of the original element into a different element, with an altered place in the periodic table. Alpha decay is a common process in heavy elements, such as uranium and radium, as they attempt to shed mass to achieve a more stable state.

Radium-226

Radium-226 (\(^ {226}_{88} Ra\)) is a radioactive isotope of radium, characterized by its instability. It naturally undergoes alpha decay to become more stable.
This decay process is a transformation that results in the radium nucleus ejecting an alpha particle, leading to the formation of a new element. The original atomic number of radium is 88, positioning it as an alkaline earth metal in the periodic table.
Upon losing an alpha particle, radium-226 (*decreases its mass number from 226 to 222* ) and (*its atomic number decreases from 88 to 86*), changing into radon-222.

• This transformation reduces radium into a more stable form.
• Radon is the element that subsequently forms, identified by the atomic number 86.

These changes are substantial because they demonstrate how elements can transform into different entities through nuclear reactions.

Nuclear Equation Balancing

Balancing nuclear equations is crucial to ensure that the transformation during nuclear reactions adheres to the law of conservation of mass and charge. Each nuclear equation must have equal sums of mass numbers and atomic numbers on both sides.
This ensures no loss of mass or electric charge occurs in the reaction. Let’s consider the generic form of a nuclear reaction involving an alpha decay:
\(^ {A}_{Z} \text{Initial Nucleus} \rightarrow ^{A-4}_{Z-2} \text{New Nucleus} + ^{4}_{2} He\)

• **Mass number balance:** The sum of mass numbers before and after the reaction must be identical.
• **Atomic number balance:** The sum of atomic numbers before and after the reaction must remain unchanged.

For example, in the decay of radium-226:
The equation \(^{226}_{88} Ra \rightarrow ^{222}_{86} Rn + ^{4}_{2} He \) showcases that both mass (226 = 222 + 4) and atomic numbers (88 = 86 + 2) remain consistent.
As evident from this equation, understanding and applying these conservation rules are key for mastering the balancing of nuclear equations, aiding in determining product elements in such reactions.