Is L decidable or not
The decision problem that corresponds to $L$ deals with the behaviour of Turing machines, not with strings of $1$'s and $0$'s.

More precisely, the decision problem is


  Given a Turing machine $M$, is it the case that $L(M) = \{ 1^n0^n \mid
&gt; n \geq 0 \}$


A decider for this decision problem would take as input the description $\langle M \rangle$ of an arbitrary Turing machine $M$ and tell us if $M$ accepted exactly those strings that are of the form $1^i0^i$ for some $i \geq 0$. A pushdown automaton that recognizes the language $\{ 1^n0^n \mid
&gt; n \geq 0 \}$ is clearly not such a decider.

A reduction from $A_{TM}$ to $L$ will, given an instance $\langle M,w \rangle$ build a description $\langle M' \rangle$ such that 

$$M \text{ accepts } w \iff L(M') = \{ 1^n0^n \mid
&gt; n \geq 0 \}$$

Here is $M'$:


  On input $x$
  
  
  If $x \neq 1^i0^i$ for some $i \geq 0$, reject
  Else simulate $M$ on $w$ and answer what $M$ answered.
  


Clearly, if $M$ accepts $w$ then $L(M') = L$. If $M$ does not accept $w$, then $L(M') = \emptyset$.