Answer
See the detailed answer below.
Work Step by Step
Since the author told us to use the conservation of momentum and the conservation of energy, we can assume that the initial $\rm ^{226}_{88}Ra$ were stationary. And since we assume that the system is isolated, then the daughter nucleus will have a net momentum of zero.
From all the above, we can use the nonrelativistic kinetic energy and momentum laws.
$$p_\alpha=p_D\tag 1$$
So,
$$KE=\frac{1}{2}mv^2$$
multiplying both sides by $2m$;
$$ 2KE m =m^2v^2=p^2\tag 2$$
And hence, the kinetic energy of the daughter nucleus is given by
$$KE_D=\frac{1}{2}m_Dv_D^2=\frac{1}{2} p_Dv_D$$
where $m_Dv_D=p_D$
$$KE_D=\frac{1}{2} p_Dv_D\times \dfrac{m_D}{m_D}=\dfrac{p_D^2}{2m_D}$$
Plugging from (1);
$$KE_D =\dfrac{p_\alpha^2}{2m_D}$$
Plugging from (2);
$$KE_D=\dfrac{ {\color{red}{\bf\not}2m_\alpha KE_\alpha} }{\color{red}{\bf\not}2m_D}$$
$$KE_D=\dfrac{ { m_\alpha } }{ m_D}KE_\alpha=\dfrac{A_\alpha}{A_D}KE_\alpha$$
where we know that the mass number of $\alpha$ is 4.
$$KE_D= \dfrac{4}{A_D}KE_\alpha\tag 3$$
The fractional of the total energy available is given by
$$ \dfrac{KE_D}{KE_\alpha+KE_D}$$
Plugging from (3);
$$\dfrac{KE_D}{KE_\alpha+KE_D}=\dfrac{\dfrac{4}{A_D}\color{red}{\bf\not}KE_\alpha}{\color{red}{\bf\not}KE_\alpha+\dfrac{4}{A_D}\color{red}{\bf\not}KE_\alpha}=\dfrac{\dfrac{4}{A_D}}{1+\dfrac{4}{A_D}}$$
Therefore,
$$\boxed{\dfrac{KE_D}{KE_\alpha+KE_D} =\dfrac{1}{\dfrac{A_D}{4}+1}}$$
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b)
We know that the $\rm ^{226}_{88}Ra$ will decay to some a daughter element of $_D=222$.
Plugging into the boxed formula above.
$$ \dfrac{KE_D}{KE_\alpha+KE_D} =\dfrac{1}{\dfrac{222}{4}+1} =0.0177$$
And hence, the energy carried away by $\alpha$ is about $$E_{away}=1-0.0177=0.982$$
which is about $\color{red}{\bf98.2}\%$ of the energy
