`._(7)^(7)Li + _(1)^(1)P rarr X`, Identify X if reaction is `(p,alpha)` type.

Options

A. `._(4)^(8)Be`
B. `._(2)^(4)He`
C. `._(0)^(0)gamma`
D. none of these

Answer

To solve the problem, we need to identify the product \( X \) in the reaction:

\[ _{7}^{7}Li + _{1}^{1}p \rightarrow X \]

where the reaction is of the type (p, alpha), meaning a proton interacts with lithium to produce an alpha particle.

### Step 1: Understand the Reaction
In this reaction, lithium-7 (\( _{7}^{7}Li \)) is reacting with a proton (\( _{1}^{1}p \)). The (p, alpha) reaction indicates that a proton is absorbed by lithium and an alpha particle is emitted.

### Step 2: Identify the Alpha Particle
An alpha particle is equivalent to a helium nucleus, which has:
- Mass number = 4
- Atomic number = 2

So, we can denote the alpha particle as \( _{2}^{4}He \).

### Step 3: Apply Conservation of Mass and Atomic Numbers
According to the conservation laws:
- The total mass number before the reaction must equal the total mass number after the reaction.
- The total atomic number before the reaction must equal the total atomic number after the reaction.

#### Mass Numbers:
Before the reaction:
- Mass number of lithium = 7
- Mass number of proton = 1
- Total mass number = \( 7 + 1 = 8 \)

After the reaction:
- Mass number of \( X \) + Mass number of alpha particle = 8
- Mass number of alpha particle = 4

Let the mass number of \( X \) be \( Y \):
\[ Y + 4 = 8 \implies Y = 4 \]

#### Atomic Numbers:
Before the reaction:
- Atomic number of lithium = 3
- Atomic number of proton = 1
- Total atomic number = \( 3 + 1 = 4 \)

After the reaction:
- Atomic number of \( X \) + Atomic number of alpha particle = 4
- Atomic number of alpha particle = 2

Let the atomic number of \( X \) be \( Z \):
\[ Z + 2 = 4 \implies Z = 2 \]

### Step 4: Identify \( X \)
Now we have:
- Mass number of \( X \) = 4
- Atomic number of \( X \) = 2

This corresponds to the element helium (\( _{2}^{4}He \)). Thus, \( X \) is helium.

### Final Answer
\[ X = _{2}^{4}He \]