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.
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 $8.96 \times 10^{-7} \ m$.
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 $8.96 \times 10^{-7} \ m$.
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