In compound `X(n, alpha) rarr ._(3)Li^(7)`, the element `X` is

Options

A. `._(2)He^(4)`
B. `._(5)B^(10)`
C. `._(5)B^(9)`
D. `._(4)Be^(11)`

Answer

To determine the element \( X \) in the reaction \( X(n, \alpha) \rightarrow ^{7}_{3}\text{Li} \), we will follow these steps:

### Step 1: Identify the Reaction Components
The reaction involves an unknown element \( X \) that undergoes a nuclear reaction with a neutron \( n \) and produces an alpha particle \( \alpha \) (which is \( ^{4}_{2}\text{He} \)) and lithium \( ^{7}_{3}\text{Li} \).

### Step 2: Write the Reaction in Terms of Atomic Mass and Atomic Number
The general form of the reaction can be written as:
\[ X + n \rightarrow ^{4}_{2}\text{He} + ^{7}_{3}\text{Li} \]
Let the atomic mass of \( X \) be \( A \) and its atomic number be \( Z \). The neutron has an atomic mass of 1 and an atomic number of 0.

### Step 3: Balance the Atomic Mass
The total atomic mass before the reaction must equal the total atomic mass after the reaction. Thus, we have:
\[ A + 1 = 4 + 7 \]
Solving this gives:
\[ A + 1 = 11 \implies A = 10 \]

### Step 4: Balance the Atomic Number
Next, we balance the atomic numbers:
\[ Z + 0 = 2 + 3 \]
Solving this gives:
\[ Z = 5 \]

### Step 5: Identify the Element
Now we have determined that the atomic mass \( A = 10 \) and the atomic number \( Z = 5 \). The element with atomic number 5 is boron (B), and specifically, the isotope is \( ^{10}_{5}\text{B} \).

### Conclusion
Therefore, the element \( X \) is \( ^{10}_{5}\text{B} \) (Boron).

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