`._(92)U^(238)` absorbs a neutron. The product emits an electron. This product further emits an electron. The result is

Options

A. `._(94)Pu^(239)`
B. `._(90)Pu^(239)`
C. `._(93)Pu^(237)`
D. .`_(94)Pu^(237)`

Explanation

To solve the problem step by step, we will analyze the decay process of Uranium-238 when it absorbs a neutron and subsequently emits electrons.

### Step 1: Identify the initial nucleus
The initial nucleus is Uranium-238, denoted as \( _{92}^{238}U \).

### Step 2: Absorption of a neutron
When Uranium-238 absorbs a neutron, its mass number increases by 1 (since a neutron is added) while the atomic number remains the same.

- New mass number: \( 238 + 1 = 239 \)
- Atomic number remains: \( 92 \)

This results in the formation of \( _{92}^{239}U \).

### Step 3: First emission of an electron
The product \( _{92}^{239}U \) undergoes beta decay, emitting an electron (beta particle). In beta decay, a neutron is converted into a proton, which increases the atomic number by 1 while the mass number remains unchanged.

- New mass number: \( 239 \) (remains the same)
- New atomic number: \( 92 + 1 = 93 \)

This results in the formation of \( _{93}^{239}Np \) (Neptunium-239).

### Step 4: Second emission of an electron
Next, the Neptunium-239 nucleus \( _{93}^{239}Np \) also undergoes beta decay, emitting another electron. Again, a neutron is converted into a proton.

- New mass number: \( 239 \) (remains the same)
- New atomic number: \( 93 + 1 = 94 \)

This results in the formation of \( _{94}^{239}Pu \) (Plutonium-239).

### Conclusion
After both emissions of electrons, the final product is \( _{94}^{239}Pu \).

### Final Result
The result of the process is Plutonium-239, denoted as \( _{94}^{239}Pu \).

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