The de-Broglie wavelength associated with an electron is 0.30 nm. Calculate: (a) the speed, and (b) the kinetic energy (in eV), of the electron.

b)a) $\displaystyle \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$

$$\mathrm{v}=\frac{\mathrm{h}}{\mathrm{~m} \lambda}=\frac{$\displaystyle 6.63$ \times $\displaystyle 10$^{-$\displaystyle 34$}}{$\displaystyle 9.1$ \times $\displaystyle 10$^{-$\displaystyle 31$} \times $\displaystyle 0.30$ \times $\displaystyle 10$^{-$\displaystyle 9$}}

$$$$ =$\displaystyle 2.41$ \times $\displaystyle 10$^{$\displaystyle 6$} \mathrm{~m} / \mathrm{s}

$$b) K.E. $\displaystyle =\frac{1}{2} m v^{2}$

$$=\frac{1}{2} \times $\displaystyle 9.1$ \times $\displaystyle 10$^{-$\displaystyle 31$} \times\left($\displaystyle 2.41$ \times $\displaystyle 10$^{$\displaystyle 6$}\right)^{$\displaystyle 2$} \mathrm{~J}

$$$$ =$\displaystyle 26.42$ \times $\displaystyle 10$^{-$\displaystyle 19$} \mathrm{~J}

$$K.E in eV

$$=\frac{$\displaystyle 26.42$ \times $\displaystyle 10$^{-$\displaystyle 19$}}{$\displaystyle 1.6$ \times $\displaystyle 10$^{$\displaystyle 19$}}=$\displaystyle 16.5$ \mathrm{eV}