A radio-active elements emits one `alpha` and `beta` particles then mass no. of daughter element is :

Options

A. Decreased by 4
B. Increased by 4
C. Decreased by 2
D. Increased by 2

Answer

To solve the problem of determining the mass number of the daughter element after a radioactive element emits one alpha particle and one beta particle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Element**:
Let the initial radioactive element be represented as \( A \) with a mass number \( m \) and atomic number \( p \).

2. **Alpha Particle Emission**:
When an alpha particle (which consists of 2 protons and 2 neutrons, with a mass number of 4) is emitted, the new element will have:
- Mass number decreases by 4 (because 2 protons and 2 neutrons are lost).
- Atomic number decreases by 2 (because 2 protons are lost).

Therefore, after the emission of one alpha particle, the new element can be represented as:
\[ \text{New mass number} = m - 4 \]
\[ \text{New atomic number} = p - 2 \]

3. **Beta Particle Emission**:
Next, when a beta particle (which is essentially an electron) is emitted, a neutron in the nucleus is converted into a proton. This process does not change the mass number but increases the atomic number by 1.

Therefore, after the emission of one beta particle, the new element will have:
- Mass number remains the same: \( m - 4 \)
- Atomic number increases by 1: \( p - 2 + 1 = p - 1 \)

4. **Final Mass Number**:
The final mass number of the daughter element after both emissions (one alpha and one beta) is:
\[ \text{Final mass number} = m - 4 \]

### Conclusion:
Thus, the mass number of the daughter element after the emissions is \( m - 4 \).