= Solution
\b[The stipulated decay branch gives $K_\alpha\simeq4.51\ \mathrm{MeV}$.] Using the supplied atomic masses, the <mass defect> is
$$
\Delta m=209.98287-205.97447-4.00260=0.00580\ \mathrm u,
$$
so the total <Q value> is $Q\simeq0.00580(931.494)=5.4027\ \mathrm{MeV}$. The photon carries away the stated $0.802\ \mathrm{MeV}$, leaving
$$
K_{\rm products}=5.4027-0.802=4.6007\ \mathrm{MeV}.
$$
Applying the recoil partition,
$$
\boxed{K_\alpha=\frac{4.6007}{1+4.00260/205.97447}=4.5130\ \mathrm{MeV}.}
$$
The daughter recoil takes approximately $0.0877\ \mathrm{MeV}$. Atomic electron counts balance in the neutral-atom mass difference; small electronic binding corrections and photon recoil are neglected consistently with the problem. These values follow the branch specified in the examination, rather than an assumed average over every physical decay branch.
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