2. If $\sin \alpha+\cos \alpha=\frac{1}{5}$, then $|\sin \alpha-\cos \alpha|=$ $\qquad$
2. $\frac{7}{5}$.

From the condition, we get
$$
\begin{array}{l}
2 \sin \alpha \cdot \cos \alpha=(\sin \alpha+\cos \alpha)^{2}-1=-\frac{24}{25} \\
\Rightarrow(\sin \alpha-\cos \alpha)^{2}=1-2 \sin \alpha \cdot \cos \alpha=\frac{49}{25} \\
\Rightarrow|\sin \alpha-\cos \alpha|=\frac{7}{5} .
\end{array}
$$