Consider the nuclear decay process: `""_(39)^(90)Yto""_(38)^(90)Sr+`? What is (are) the missing product (s)?

Options

A. A photon
B. A positron and a neutrino
C. A proton
D. An electron and an alpha particle

Answer

To solve the nuclear decay process \( _{39}^{90}Y \to _{38}^{90}Sr + ? \), we need to identify the missing product(s) in this decay reaction.

### Step-by-Step Solution:

1. **Identify the Decay Process**:
- We are given the decay of Yttrium-90 (\( _{39}^{90}Y \)) into Strontium-90 (\( _{38}^{90}Sr \)).
- The general form of a nuclear decay can be represented as:
\[ _Z^A X \to _{Z'}^{A'} Y + ? \]
- Here, \( Z \) is the atomic number (number of protons), \( A \) is the mass number (total number of protons and neutrons), \( X \) is the parent nucleus, and \( Y \) is the daughter nucleus.

2. **Determine Changes in Atomic and Mass Numbers**:
- In this decay, the atomic number decreases from 39 (Yttrium) to 38 (Strontium), indicating that a particle with a positive charge is emitted.
- The mass number remains the same (90), which means that the total number of nucleons (protons + neutrons) is conserved.

3. **Identify the Missing Particle**:
- Since the atomic number decreases by 1, we can conclude that a positron (\( e^+ \)) is emitted. A positron is a particle with the same mass as an electron but with a positive charge.
- The emission of a positron can be represented as:
\[ e^+ \]
- Additionally, in beta decay processes where a positron is emitted, a neutrino (\( \nu \)) is also released. This is a neutral particle that carries away energy.

4. **Write the Complete Decay Equation**:
- Therefore, the complete decay process can be written as:
\[ _{39}^{90}Y \to _{38}^{90}Sr + e^+ + \nu \]
- Here, \( e^+ \) is the positron and \( \nu \) is the neutrino.

### Final Answer:
The missing products in the decay process are a positron (\( e^+ \)) and a neutrino (\( \nu \)).

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