An element `""_(96)X^(227)` emits `4alpha` and `5beta` particles to form new element Y. Then atomic number and mass number of Y are

Options

A. 93, 211
B. 211, 93
C. 212, 88
D. 88, 211

Explanation

To solve the problem of determining the atomic number and mass number of the new element Y formed after the emission of 4 alpha and 5 beta particles from the element \( _{96}X^{227} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Atomic and Mass Numbers:**
- The element \( X \) has an atomic number (Z) of 96 and a mass number (A) of 227.
- Therefore, we have:
- \( Z_X = 96 \)
- \( A_X = 227 \)

2. **Calculate the Effect of Alpha Particle Emission:**
- Each alpha particle emission decreases the atomic number by 2 and the mass number by 4.
- For 4 alpha particles:
- Change in atomic number = \( 4 \times 2 = 8 \)
- Change in mass number = \( 4 \times 4 = 16 \)

3. **Calculate the Effect of Beta Particle Emission:**
- Each beta particle emission increases the atomic number by 1 and does not change the mass number.
- For 5 beta particles:
- Change in atomic number = \( 5 \times 1 = 5 \)
- Change in mass number = \( 5 \times 0 = 0 \)

4. **Calculate the Final Atomic Number of Element Y:**
- The total change in atomic number:
\[ Z_Y = Z_X - \text{(change due to alpha)} + \text{(change due to beta)} \]
\[ Z_Y = 96 - 8 + 5 = 93 \]

5. **Calculate the Final Mass Number of Element Y:**
- The total change in mass number:
\[ A_Y = A_X - \text{(change due to alpha)} + \text{(change due to beta)} \]
\[ A_Y = 227 - 16 + 0 = 211 \]

6. **Final Result:**
- The atomic number of element Y is 93.
- The mass number of element Y is 211.

### Conclusion:
The atomic number and mass number of the new element Y are:
- Atomic Number (Z) = 93
- Mass Number (A) = 211