Answer
$2-\frac{2}{e}, $ the integral converges.
Work Step by Step
Let $ y=\sqrt{x},\ \ dy=\frac{dx}{2\sqrt{x}}$
\begin{align*}
\int_{0}^{1} \frac{e^{-\sqrt{x}}}{\sqrt{x}} dx&= 2\int_{0}^{1} e^{-y} d y\\
&=-e^{-y}\bigg|_0^1 \\
&=2-\frac{2}{e},
\end{align*}
Then the integral converges.