Question

Calculate the kinetic energy of the electron having wavelength 1 nm.

• A. 2.1 eV
• B. 3.1 eV
• C. 1.5 eV
• D. 4.2 eV

Answer 1

Answer: (C)

de broglie wavelength,
$\lambda =\frac{h}{\sqrt{2mk}}$
or $k=\frac{h^2}{2m{\lambda }^2}$
$=\frac{{\left(6.67\times 10^{-34}\right)}^2}{2\times 9.1\times 10^{-31}{\left(1\times 10^{-9}\right)}^2}=1.5eV.$