In the given nuclear reaction `A, B, C, D, E` represents `._92 U^238 rarr^(alpha) ._BTh^A rarr^(beta) ._D Pa^C rarr^(E) ._92 U^234`.

Options

A. `A=234,B=90,C=234,D=91andE=beta`
B. `A=234,B=90,C=238,D=94andE=alpha`
C. `A=238,B=93,C=234,D=91andE=beta`
D. `A=234,B=90,C=234,D=93andE=alpha`

Explanation

To solve the given nuclear reaction step by step, we will analyze each part of the reaction and determine the values of A, B, C, D, and E.

### Step 1: Identify the initial reaction
The initial reaction is:
\[ _{92}^{238}U \rightarrow \text{(alpha particle)} + _{B}^{A}Th \]
Here, Uranium-238 (U-238) undergoes alpha decay.

### Step 2: Determine the products of alpha decay
An alpha particle consists of 2 protons and 2 neutrons, which means it has a mass number of 4 and an atomic number of 2. When U-238 emits an alpha particle, the new element formed will have:
- Mass number: \(238 - 4 = 234\)
- Atomic number: \(92 - 2 = 90\)

Thus, we have:
\[ A = 234, \quad B = 90 \]
The resulting element is Thorium (Th).

### Step 3: Write the first decay product
The first decay product is:
\[ _{90}^{234}Th \rightarrow \text{(beta particle)} + _{D}^{C}Pa \]
Here, Thorium-234 undergoes beta decay.

### Step 4: Determine the products of beta decay
In beta decay, a neutron is converted into a proton, resulting in:
- Mass number remains the same: \(234\)
- Atomic number increases by 1: \(90 + 1 = 91\)

Thus, we have:
\[ C = 234, \quad D = 91 \]
The resulting element is Protactinium (Pa).

### Step 5: Write the second decay product
The second decay product is:
\[ _{91}^{234}Pa \rightarrow \text{(E)} + _{92}^{234}U \]
Here, Protactinium-234 undergoes decay to form Uranium-234.

### Step 6: Determine the type of decay for the last reaction
In this case, since the atomic number increases from 91 to 92 while the mass number remains the same, this indicates that a beta particle is emitted.

Thus, we have:
\[ E = \text{beta particle} \]

### Final Values
Now, we can summarize the values:
- \(A = 234\)
- \(B = 90\)
- \(C = 234\)
- \(D = 91\)
- \(E = \text{beta particle}\)

### Final Answer
The values are:
- \(A = 234\)
- \(B = 90\)
- \(C = 234\)
- \(D = 91\)
- \(E = \text{beta particle}\)