Are DFAs with a unary alphabet strictly less powerful than DFAs with a binary alphabet? Are DFAs with a unary alphabet strictly less powerful than DFAs with a binary alphabet? Is this even a meaningful question?

For example, if $\Sigma = \{\texttt{0}, \texttt{1}\}$, we can encode any larger alphabet using $\Sigma$, but if $\Sigma = \{\texttt{0}\}$, this can define a DFA (that say, recognizes $L = \{ \texttt{0}^k \mid k &gt; 0\}$)... but such a DFA would never be able to recognize more "complex" regular expressions. For example, there is no way to encode $\texttt{0011}$ using a unary alphabet that a DFA would recognize (we could use, say, Godel numbering, but that would require a more powerful machine that could "count").

If DFAs with a unary alphabet less powerful than DFAs with a binary alphabet, is there a name for this language/grammar? I recognize this is kind of an odd question, since the DFA that recognizes $L = \{ \texttt{0}^k \mid k &gt; 0\}$ recognizes all unary languages... but technically there still are a countably infinite number of DFAs in this class ($L = \{ \texttt{0}^1 \}$, $L=\{\texttt{0}^2\}$, etc.)

Note I am of course assuming that for $\Sigma = \{ \texttt{0} \}$, that it does not contain the empty symbol $\varepsilon$.
Take the set $A=\{2^k \mid k \ge 1\}$. The language 
$$ L=\{ \text{bin}(a) \mid a \in A\} = \{10^k\mid k \ge 0 \}$$
is clearly a regular language, whereas 
$$L=\{ 1^a \mid a \in A\} $$
is not regular. Proving the latter is a standard textbook application of the regular pumping lemma.