Yes, you can apply master's theorem, but you won't need to when you get to the right form.
Let $n = 2^{k}$. The recurrence becomes : $T( 2^{k}) = 2^{k/2}T( 2^{k/2}) + 2^{k}$. Now divide the entire equation by $2^{k}$. We get :
$T( 2^{k}) / 2^{k}= T( 2^{k/2})/2^{k/2} + 1$ Let $F(k) = T( 2^{k}) / 2^{k}$. Thus, what we have is: $F(k) = F(k/2) + 1$.
Apply master's theorem if you want but why even bother if you know what time is taken by binary search.Cheers.