As given that, $v=v_0 \vec{i}$ and $B=B_0 \vec{j}$
Force on moving electron due to perpendicular magnetic field $B$ is, $F=-e ~(v \times B)$
$$
\begin{array}{ll}
\mathrm{F}=-\mathrm{e}\left[\mathrm{v}_0 \hat{\mathrm{i}} \times \mathrm{B}_0 \hat{\mathrm{i}}\right]=-\mathrm{ev}_0 \mathrm{~B}_0(\hat{\mathrm{i}} \times \hat{\mathrm{j}}) \\
\Rightarrow=-\mathrm{ev}_0 \mathrm{~B}_0 \hat{\mathrm{k}} & (\because \hat{\mathrm{k}}=\hat{\mathrm{i}} \times \hat{\mathrm{j}})
\end{array}
$$
So, the force is perpendicular to $\mathrm{v}$ and $\mathrm{B}$, both as the force is $\perp$ to the velocity so the magnitude of $\mathrm{v}$ will not change, so momentum is $(=\mathrm{mv})$ will remain same or constant in magnitude. Hence,
de-Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ remians constant.