de-Broglie wavelength

$$\begin{gathered} \lambda =\frac{h}{mv}\Rightarrow v=\frac{h}{m\lambda } \\ Here, h=6.6\times {10}^{-34}\mathrm{Js} \end{gathered}$$

and for electron,  $m= 9 x {10}^{-31} kg$

Now consider each option one by one

$$\begin{gathered} \left(1\right) {\lambda }_1=10\mathrm{nm}=10\times {10}^{-9}\mathrm{m}={10}^{-8}\mathrm{m} \\ \Rightarrow {\mathrm{v}}_1=\frac{6.6\times {10}^{-34}}{\left(9\times {10}^{-31}\right)\times {10}^{-8}} \\ \Rightarrow =\frac{2.2}{3}\times {10}^5\approx {10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} \text{ (2) }{\lambda }_2={10}^{-1}\mathrm{nm}={10}^{-1}\times {10}^{-9}\mathrm{m}={10}^{-10}\mathrm{m} \\ \Rightarrow {\mathrm{v}}_2=\frac{6.6\times {10}^{-34}}{\left(9\times {10}^{-31}\right)\times {10}^{-10}}\approx {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} \text{ (3) }{\lambda }_3={10}^{-4}\mathrm{nm}={10}^{-4}\times {10}^{-9}\mathrm{m}={10}^{-13}\mathrm{m} \\ \Rightarrow {\mathrm{v}}_3=\frac{6.6\times {10}^{-34}}{\left(9\times {10}^{-31}\right)\times {10}^{-13}}\approx {10}^{10}\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} \text{ (4) }{\lambda }_4={10}^{-6}\mathrm{nm}={10}^{-6}\times {10}^{-9}\mathrm{m}={10}^{-15}\mathrm{m} \\ \Rightarrow {\mathrm{v}}_4=\frac{6.6\times {10}^{-34}}{9\times {10}^{-31}\times {10}^{-15}}\approx {10}^{12}\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, options (3) and (4) are correct as v₃ and v₄ is greater
than 3 x 10⁸ m/s.