Answer
Converges.
Work Step by Step
Suppose $y=\sqrt{x} \implies dy=\dfrac{dx}{2\sqrt{x}}$
Now, $\int_{0}^{1} \dfrac{e^{-\sqrt{x}}}{\sqrt{x}} dx= 2\int_{0}^{1} e^{-y} d y$
or, $=-2e^{-y}|_0^1$
or,$=-2[e^{-1}-1]$
or, $=2-\dfrac{2}{e}$
Thus, the integral converges.