Question

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

Answer 1

The correct option is B

16.8 eV

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