Question

The kinetic energy of an electron with de-Broglie wavelength of $0.3$ nanometre is

• A. 0.168 eV
• B. 16.8 eV
• C. 1.68 eV
• D. 2.5 eV

Answer 1

Answer: (B)

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