The change in the nucleus can be calculated by applying the displacement laws for radioactive decay:

1. $\alpha$-decay: The emission of one \alpha particle ($^4_2\text{He}$) reduces the mass number ($m$) by 4 and the atomic number ($n$) by 2 . Result: $_{n-2} Y^{m-4}$.
2. $\beta$-decay: The emission of one \beta particle ($^0_{-1}e$) increases the atomic number by 1 while leaving the mass number unchanged. For two $\beta$ particles, the atomic number increases by $2 \times 1 = 2$. Result: $_{ (n-2) + 2 } X^{m-4} = _n X^{m-4}$.

Since the final atomic number ($n$) is the same as the initial one, the chemical identity of the element remains the same ($X$). Therefore, the resulting nucleus is an isotope of the parent element with a mass number of $m-4$.