C

Step 1: Let's analyze the reaction. The Boron nucleus, represented as \( \text{B} \) typically has a mass number of 10 (\( \text{^{10}B} \)).
Step 2: When it is bombarded by neutrons (\( \text{n} \)), a neutron is added to the Boron nucleus, resulting in \( \text{^{11}B} \).
Step 3: Next, the emission of an alpha particle (\( \text{\alpha} \)) occurs. An alpha particle consists of 2 protons and 2 neutrons, which means we lose 4 mass units and 2 protons from the nucleus.
Step 4: The reaction can be summarized as:
\( \text{^{11}B} + \text{n} \rightarrow \text{^{7}Be} + \text{\alpha} \)
Here, \( \text{^{7}Be} \) is the resulting nucleus.
Step 5: Therefore, the final element is Beryllium (\( \text{Be} \)) and it has a mass number of 7.
Therefore, the final answer is:
**Element: Be; Mass Number: 7.**