Answer
See the detailed answer below.
Work Step by Step
a) The author asks us to write the decay equation that gives the radon gas $\rm ^{222}_{86}Rn$ and $\rm ^4_2He$
$$\rm ^{226}_{88}Ra\rightarrow\;\;^{222}_{86}Rn+\;^4_2He$$
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b) Since we will ignore the kinetic energy of the daughter nucleus, the kinetic energy of $\alpha$ is given by
$$KE_{\rm ^4_2He}=Q_{\rm ^{226}_{88}Ra}-Q_{\rm ^{222}_{86}Rn}-Q_{\rm
^4_2He}$$
$$KE_{\rm ^4_2He}=\left(m_{\rm ^{226}_{88}Ra}-m_{\rm ^{222}_{86}Rn}-m_{\rm ^4_2He}\right)931.5$$
Plugging the known;
$$KE_{\rm ^4_2He}=\left( 226.025410- 222.017578-4.002603\right) \times 931.5$$
$$KE_{\rm ^4_2He}=\color{red}{\bf 4.871}\;\rm MeV$$
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c) Since the energy of the alpha particle is less than the threshold of relativism, we can use the momentum law.
And since the momentum is conserved, then
$$p_{\rm ^4_2He}=p_{\rm ^{222}_{86}Rn}$$
where $p=\sqrt{2mKE}$;
Thus,
$$p_{\rm ^4_2He}=p_{\rm ^{222}_{86}Rn}=\sqrt{2m_{\rm ^4_2He}KE_{\rm ^4_2He}}$$
Plugging the known;
$$p_{\rm ^4_2He}=p_{\rm ^{222}_{86}Rn}=\sqrt{2\times 4.002603 \times 4.871\times 931.5 }$$
$$p_{\rm ^4_2He}=p_{\rm ^{222}_{86}Rn}=\color{red}{\bf 190.6}\;{\rm MeV}/c$$
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d) We know that
$$KE=\dfrac{p^2}{2m}$$
Thus,
$$KE_{\rm ^{222}_{86}Rn}=\dfrac{p_{\rm ^{222}_{86}Rn}^2}{2m_{\rm ^{222}_{86}Rn}}$$
Plugging the known;
$$KE_{\rm ^{222}_{86}Rn}=\dfrac{190.6^2}{2(222.017578)(931.5)}$$
$$KE_{\rm ^{222}_{86}Rn}= \color{red}{\bf 0.0878}\;\rm MeV$$
Now it is obvious that it was valid to ignore the kinetic energy of the daughter nuclei since its energy is less than $2\%$ of the entry of particle alpha.
$$\dfrac{KE_{_{\rm ^{222}_{86}Rn}}}{KE_{\rm ^4_2He}}=\dfrac{0.0878}{4.871}$$
$${KE_{_{\rm ^{222}_{86}Rn}}}=\bf0.018\;{KE_{\rm ^4_2He}}$$