The mass of an electron is $9.1 \times 10^{-31} \ kg$. If its $K.E.$ is $3.0 \times 10^{-25} \ J$,calculate its wavelength.

(N/A) From de Broglie's equation,$\lambda = \frac{h}{mv}$.
Given,Kinetic energy $(K.E.)$ of the electron $= 3.0 \times 10^{-25} \ J$.
Since $K.E. = \frac{1}{2} mv^{2}$,therefore velocity $(v) = \sqrt{\frac{2 K.E.}{m}}$.
$v = \sqrt{\frac{2(3.0 \times 10^{-25} \ J)}{9.1 \times 10^{-31} \ kg}} = \sqrt{6.5934 \times 10^{5}} \approx 812 \ ms^{-1}$.
Substituting the value in the expression of $\lambda$:
$\lambda = \frac{6.626 \times 10^{-34} \ Js}{(9.1 \times 10^{-31} \ kg)(812 \ ms^{-1})} \approx 8.96 \times 10^{-7} \ m$.
Hence,the wavelength of the electron is $8.96 \times 10^{-7} \ m$.