B

Step 1: Identify the reaction involved when a Boron nucleus (\( _5^{11}B \)) is bombarded by a neutron (\( _0^{1}n \)).
Step 2: Write the nuclear reaction:
\[ _5^{11}B + _0^{1}n \rightarrow _{Z}^{A}X + _2^{4}He \]
Step 3: Conservation of nucleons:
According to the conservation of mass number: \( 11 + 1 = A + 4 \) \( \Rightarrow A = 8 \)
Step 4: Conservation of charge:
From the reactants' charge (5 from Boron) minus the charge of the emitted alpha particle (2), we deduce the charge of the product: \( 5 = Z + 2 \) \( \Rightarrow Z = 3 \)
Step 5: The resulting nucleus is therefore \( _3^{8}Li \) (Lithium), with a mass number of 8.
Therefore, the element is Lithium and the mass number is 8.