Write the nuclear reaction

${}_{29}^{63}\mathrm{Cu} + {}_{1}^{1}\mathrm{H} \rightarrow 6\,{}_{0}^{1}\mathrm{n} + {}_{2}^{4}\alpha + 2\,{}_{1}^{1}\mathrm{H} + X$

Use conservation of mass number

Total mass number on LHS: $63 + 1 = 64$

Total mass number on RHS excluding $X$:

• $6$ neutrons $\Rightarrow 6 \times 1 = 6$
• one $\alpha$ particle $\Rightarrow 4$
• $2$ protons $\Rightarrow 2 \times 1 = 2$

So, $A_X = 64 - (6+4+2) = 64 - 12 = 52$

Hence, mass number of $X$ is $A_X = 52$

Use conservation of atomic number

Total atomic number on LHS: $29 + 1 = 30$

Total atomic number on RHS excluding $X$:

• neutrons contribute $0$
• $\alpha$ particle contributes $2$
• $2$ protons contribute $2 \times 1 = 2$

Therefore, $Z_X = 30 - (2+2) = 30 - 4 = 26$

So element $X$ has atomic number $Z_X = 26$

Identify the element

Atomic number $26$ corresponds to iron (Fe).

Find its group

Iron belongs to Group 8 of the periodic table.

Final answer

$\boxed{8}$

Comparison with stored correct answer

Stored correct answer = $8$

This matches the derived answer.