A radioactive element A on disintegration gives two elements B and C. If B is helium and C is the element of atomic number 90 and atomic mass 234, the element A is:

Options

A. `_92^238U`
B. `_88^234Ra`
C. `_90^234Th`
D. `_91^234 Pa`

Answer

To solve the problem, we need to identify the radioactive element A based on the information provided about its disintegration products B and C.

### Step-by-Step Solution:

1. **Identify the Products of Disintegration**:
- We know that element A disintegrates to form two elements: B and C.
- Element B is specified as helium (He), which has:
- Atomic number (Z) = 2
- Atomic mass (A) = 4

2. **Analyze Element C**:
- Element C has an atomic number of 90 and an atomic mass of 234.
- This means:
- Atomic number (Z) = 90
- Atomic mass (A) = 234

3. **Calculate Total Atomic Number and Mass Number**:
- According to the law of conservation of mass and charge in nuclear reactions:
- Total atomic number of products = Atomic number of A
- Total mass number of products = Mass number of A
- For the atomic number:
- Total atomic number = Z(He) + Z(C) = 2 + 90 = 92
- For the mass number:
- Total mass number = A(He) + A(C) = 4 + 234 = 238

4. **Determine Atomic Number and Mass Number of Element A**:
- From the calculations:
- Atomic number of A = 92
- Mass number of A = 238

5. **Identify Element A**:
- The element with atomic number 92 is Uranium (U).
- Therefore, element A is Uranium with the notation:
- Uranium-238 (U-238).

### Final Answer:
The element A is Uranium (U) with atomic number 92 and atomic mass 238.

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