Question

In the nuclear reaction $\begin{smallmatrix}197\\ 80\end{smallmatrix}X \rightarrow \begin{smallmatrix}197\\ 79\end{smallmatrix}Y + Z + v$, the particle Z is ________.

• A. $\alpha$ particle
• B. $\beta^{+}$ particle
• C. $\beta^{-}$ particle
• D. proton
• E. neutron

Hint:

$\beta^{+}$ decay: Atomic number decreases by 1; $\beta^{-}$ decay: Atomic number increases by 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This question requires us to identify an unknown particle (Z) in a nuclear reaction. To do this, we must apply the laws of conservation of mass number (the superscript) and atomic number (the subscript, which represents charge).
Step 2: Key Formula or Approach
In any nuclear reaction of the form \( {}^{A_1}_{Z_1}\text{P} \to {}^{A_2}_{Z_2}\text{D} + {}^{A_3}_{Z_3}\text{E} \), the following must hold:
- Conservation of Mass Number: \(A_1 = A_2 + A_3\)
- Conservation of Atomic Number (Charge): \(Z_1 = Z_2 + Z_3\)
The symbol \(\nu\) represents a neutrino, which has a mass number of 0 and an atomic number of 0.
Step 3: Detailed Explanation
1. Analyze the given reaction.
\[ {}^{197}_{80}\text{X} \to {}^{197}_{79}\text{Y} + {}^{A}_{Z}\text{Z} + \nu \] 2. Apply Conservation of Mass Number.
The mass number on the left is 197. The mass numbers on the right are 197 for Y, A for Z, and 0 for \(\nu\).
\[ 197 = 197 + A + 0 \] \[ A = 0 \] So, particle Z has a mass number of 0. This eliminates \(\alpha\) particle (\({}^4_2\text{He}\)), proton (\({}^1_1\text{p}\)), and neutron (\({}^1_0\text{n}\)). The remaining possibilities are electron-like particles.
3. Apply Conservation of Atomic Number (Charge).
The atomic number on the left is 80. The atomic numbers on the right are 79 for Y, Z for particle Z, and 0 for \(\nu\).
\[ 80 = 79 + Z + 0 \] \[ Z = 80 - 79 = 1 \] So, particle Z has an atomic number (charge) of +1.
4. Identify Particle Z.
We are looking for a particle with mass number \(A=0\) and charge \(Z=+1\).
- A \(\beta^-\) particle (electron) is denoted as \({}^0_{-1}e\).
- A \(\beta^+\) particle (positron) is denoted as \({}^0_{+1}e\).
Our particle Z has \(A=0\) and \(Z=+1\), which corresponds to a positron, or \(\beta^+\) particle.
This type of decay, where a proton in the nucleus turns into a neutron while emitting a positron and a neutrino (\( p \to n + e^+ + \nu \)), is called positron emission or \(\beta^+\) decay. It occurs in proton-rich nuclei.
Step 4: Final Answer
The particle Z is a \(\beta^+\) particle (positron).