The nucleus resulting from `""_(92)U^(238)` after successive emission of two `alpha` and four `beta` -particles is

Options

A. `""_(90)Th^(230)`
B. `""_(94)Pu^(230)`
C. `""_(88)Ra^(230)`
D. `""_(92)U^(230)`

Explanation

To solve the problem of determining the nucleus resulting from the successive emission of two alpha particles and four beta particles from uranium-238, we can follow these steps:

### Step 1: Understand the Initial Nucleus
The initial nucleus is uranium-238, which has:
- Atomic number (Z) = 92 (number of protons)
- Mass number (A) = 238 (total number of protons and neutrons)

### Step 2: Calculate the Effect of Alpha Emission
Alpha particles are equivalent to helium nuclei, which means:
- Each alpha emission decreases the mass number by 4 and the atomic number by 2.

Since we have 2 alpha emissions:
- Mass number after 2 alpha emissions = 238 - 2(4) = 238 - 8 = 230
- Atomic number after 2 alpha emissions = 92 - 2(2) = 92 - 4 = 88

### Step 3: Calculate the Effect of Beta Emission
Beta particles (specifically beta minus) do not change the mass number but increase the atomic number by 1 for each emission.

Since we have 4 beta emissions:
- Mass number remains the same = 230
- Atomic number after 4 beta emissions = 88 + 4 = 92

### Step 4: Determine the Final Nucleus
After performing the calculations:
- Final mass number = 230
- Final atomic number = 92

Now, we need to identify the element corresponding to atomic number 92 and mass number 230. The element with atomic number 92 is uranium (U).

### Final Answer
The nucleus resulting from the successive emission of two alpha and four beta particles from uranium-238 is **Uranium-230** (U-230).

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