A radioactive element X emits six `alpha`-particles and four `beta`-particles leading to end product `""_(82)^(208)Pb`. X is

Options

A. `""_(92)^(238)U`
B. `""_(90)^(230)Th`
C. `""_(90)^(232)Th`
D. `""_(92)^(239)U`

Answer

To determine the radioactive element X that emits six alpha particles and four beta particles, leading to the end product of lead (Pb) with atomic number 82 and mass number 208, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Alpha and Beta Particles**:
- An alpha particle (α) consists of 2 protons and 2 neutrons. Therefore, it has an atomic number of 2 and a mass number of 4.
- A beta particle (β) is essentially an electron, which has negligible mass and a charge of -1.

2. **Initial Variables**:
- Let the atomic number of element X be \( Z \) and its mass number be \( A \).

3. **Changes in Atomic Mass**:
- When 6 alpha particles are emitted, the total mass loss due to alpha emission is \( 6 \times 4 = 24 \) (since each alpha particle has a mass number of 4).
- Therefore, the final mass number after the emission will be:
\[ A - 24 = 208 \quad \text{(mass number of lead)} \]
- Solving for \( A \):
\[ A = 208 + 24 = 232 \]

4. **Changes in Atomic Number**:
- The atomic number decreases by 6 due to the emission of 6 alpha particles (each alpha particle reduces the atomic number by 2).
- The atomic number increases by 4 due to the emission of 4 beta particles (each beta particle increases the atomic number by 1).
- Therefore, the final atomic number after all emissions will be:
\[ Z - 6 + 4 = 82 \quad \text{(atomic number of lead)} \]
- Simplifying gives:
\[ Z - 2 = 82 \implies Z = 84 \]

5. **Identifying Element X**:
- Now we have the atomic number \( Z = 84 \) and mass number \( A = 232 \).
- The element with atomic number 84 is Polonium (Po).

### Conclusion:
The radioactive element X is Polonium (Po).