The radioactive decay of `""_(35)X^(88)` by a `beta`-emission produces an unstable nucleus which spontaneously emits a neutron. The final product is

Options

A. `""_(37)X^(88)`
B. `""_(35)X^(89)`
C. `""_(34)Z^(38)`
D. `""_(36)W^(87)`

Explanation

To solve the problem, we need to analyze the radioactive decay process step by step.

### Step 1: Identify the initial nucleus
The initial nucleus is given as \( _{35}X^{88} \). Here, \( 35 \) is the atomic number (Z) and \( 88 \) is the mass number (A).

### Step 2: Understand beta emission
In beta emission, a neutron in the nucleus is converted into a proton, and a beta particle (which is an electron) is emitted. This process increases the atomic number by 1 while the mass number remains unchanged.

### Step 3: Calculate the new atomic number and mass number after beta emission
After beta emission:
- The new atomic number = \( 35 + 1 = 36 \)
- The mass number remains the same = \( 88 \)

Thus, the nucleus after beta emission will be \( _{36}Y^{88} \) (where Y is the new element).

### Step 4: Identify the unstable nucleus
The nucleus \( _{36}Y^{88} \) is unstable and spontaneously emits a neutron.

### Step 5: Calculate the new atomic number and mass number after neutron emission
When a neutron is emitted:
- The atomic number remains the same = \( 36 \)
- The mass number decreases by 1 = \( 88 - 1 = 87 \)

Thus, the final product after neutron emission will be \( _{36}Y^{87} \).

### Step 6: Identify the final product
The final product is \( _{36}Y^{87} \).

### Conclusion
The final product of the decay process is \( _{36}Y^{87} \).

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