A radium `Ra_88^224` isotope, on emission of an `alpha`-particle gives rise to a new element whose mass number and atomic number will be:

Options

A. 220 and 86
B. 225 and 87
C. 228 and 88
D. 224 and 86

Answer

To solve the problem of determining the mass number and atomic number of the new element formed when the radium isotope \( Ra_{88}^{224} \) emits an alpha particle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Isotope**:
The given isotope is radium, represented as \( Ra_{88}^{224} \). Here, the atomic number (Z) is 88 and the mass number (A) is 224.

2. **Understand Alpha Particle Emission**:
An alpha particle is essentially a helium nucleus, which has a mass number of 4 and an atomic number of 2. This can be represented as \( \alpha_{2}^{4} \).

3. **Set Up the Nuclear Reaction**:
When radium emits an alpha particle, the reaction can be represented as:
\[ Ra_{88}^{224} \rightarrow X_{Z}^{A} + \alpha_{2}^{4} \]
where \( X \) is the new element formed, and we need to find its mass number \( A \) and atomic number \( Z \).

4. **Conservation of Mass Number**:
According to the law of conservation of mass number:
\[ 224 = A + 4 \]
Rearranging gives:
\[ A = 224 - 4 = 220 \]

5. **Conservation of Atomic Number**:
Similarly, for atomic numbers:
\[ 88 = Z + 2 \]
Rearranging gives:
\[ Z = 88 - 2 = 86 \]

6. **Conclusion**:
The new element \( X \) formed after the emission of the alpha particle has:
- Mass number \( A = 220 \)
- Atomic number \( Z = 86 \)

### Final Answer:
The new element formed has a mass number of 220 and an atomic number of 86.