• # question_answer The kinetic energy of an electron with de-Broglie wavelength of 0.3 nanometer is         A) 0.168 eV B) 16.8 eV C) 1.68 eV D) 2.5 eV

[b] $\lambda =\frac{h}{\sqrt{2mE}}\Rightarrow E=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}$ $=\frac{{{(6.6\times {{10}^{-34}})}^{2}}}{2\times 9.1\times {{10}^{-31}}\times {{(0.3\times {{10}^{-9}})}^{2}}}=2.65\times {{10}^{-18}}J$ $=16.8\,eV$