In the disintegration series
`._(92)^(238)U overset rarr (alpha) X overset rarr (beta^(-))._Z^(A)Y` the values of `Z` and `A`, respectively, will be

Options

A. `92` , `326`
B. `88` , `230`
C. `90` , `234`
D. `91` , `234`

Explanation

To solve the problem, we need to analyze the disintegration series of Uranium-238 (U-238) as it undergoes alpha and beta decay.

### Step 1: Understand the Decay Process
Uranium-238 (represented as \( _{92}^{238}U \)) undergoes two types of decay: alpha decay followed by beta decay.

### Step 2: Alpha Decay
In alpha decay, the nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons. This results in:
- A decrease in mass number (A) by 4 (since 2 protons + 2 neutrons = 4).
- A decrease in atomic number (Z) by 2 (since 2 protons are lost).

Calculating the new values after alpha decay:
- New mass number \( A = 238 - 4 = 234 \)
- New atomic number \( Z = 92 - 2 = 90 \)

So after alpha decay, we have:
\[ _{90}^{234}X \]

### Step 3: Beta Decay
In beta decay, a neutron is converted into a proton, and a beta particle (an electron) is emitted. This results in:
- No change in mass number (A remains the same).
- An increase in atomic number (Z) by 1 (since one proton is added).

Calculating the new values after beta decay:
- Mass number remains \( A = 234 \)
- Atomic number \( Z = 90 + 1 = 91 \)

So after beta decay, we have:
\[ _{91}^{234}Y \]

### Final Values
Thus, the final values of \( Z \) and \( A \) for the element \( Y \) are:
- \( Z = 91 \)
- \( A = 234 \)

### Conclusion
The values of \( Z \) and \( A \) respectively will be \( 91 \) and \( 234 \).

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