= Solution
For a parent at rest, <momentum conservation> gives equal and opposite daughter and <alpha particle> <momenta>. In the nonrelativistic recoil approximation,
$$
Q=T_\alpha+T_D=\frac{p^2}{2m_\alpha}+\frac{p^2}{2M_D},\qquad
\boxed{T_\alpha=Q\frac{M_D}{M_D+m_\alpha}.}
$$
Approximating nuclear <masses> by their <mass numbers>, $M_D\simeq(A-4)m_u$ and $m_\alpha\simeq4m_u$, gives
$$
\boxed{T_\alpha\simeq\frac{A-4}{A}Q.}
$$
The printed phrase “atomic number $A$” must here mean \b[<mass number>]. <Atomic number> alone does not determine the recoil fraction. For example, ${}^{210}\mathrm{Po}$ and ${}^{212}\mathrm{Po}$ have the same <atomic number> $84$, but the approximate fractions are $206/210$ and $208/212$, respectively.
For completeness, exact <relativistic two-body decay> <kinematics> gives
$$
T_\alpha=\left(\frac{M_P^2+m_\alpha^2-M_D^2}{2M_P}-m_\alpha\right)c^2
=Q\frac{M_D+Q/(2c^2)}{M_P},
$$
where $M_P=M_D+m_\alpha+Q/c^2$. Expanding for $Q\ll M_Pc^2$ recovers the recoil-sharing expression.
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