(a) Work done by field equals KE gained: $qV=\dfrac12mv^2 \Rightarrow v=\sqrt{\dfrac{2qV}{m}}$

de Broglie wavelength:

$$\lambda=\frac{h}{mv}=\frac{h}{\sqrt{2mqV}}$$

(b) Since $\lambda \propto \dfrac{1}{\sqrt V}$, a graph of $\lambda$ vs $V$ is a decreasing curve (like $1/\sqrt{x}$) approaching zero as $V$ increases; a graph of $\lambda$ vs $\dfrac{1}{\sqrt V}$ would instead be a straight line through the origin.