The mass of electron is $9.1\times {10}^{-31} \mathrm{kg}.$ If the kinetic energy of electron is $3.0\times {10}^{-25} \mathrm{J}.$ Determine the wavelength (in nm).

de-Broglie equation in terms of kinetic energy is

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{m}\left(\mathrm{KE}\right)}} \\ =\frac{6.626\times {10}^{-34 }\mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-1}}{\sqrt{2\times 9.1\times {10}^{-31} \mathrm{kg}\times 3.0\times {10}^{-25} \mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-1}}} \\ =8.967\times {10}^{-7}\mathrm{m} \\ =896.7\times {10}^{-9}\mathrm{m} \\ =896.7\mathrm{nm} \end{gathered}$$