Question

An electron has de Broglie wavelength 2.79×10−10 m. Determine the electron's kinetic energy in electron volts.

Answer 1

Since the formula of de Broglie wavelength,

$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2Km}}$ ...........................(1)

where K is the kinetic energy of electron.

given that the de Broglie wavelength $\lambda = 2.79 \times 10^{-10}$ m

hence the kinetic energy of electron by equation (1) can be written as,

$K = \frac{h^2}{2m \lambda^2}$

$K = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (2.79 \times 10^{-10})^2}$

$K = 3.09561 \times 10^{-18}$ Joules.

hence the kinetic energy in electron volts is that,

$K = \frac{3.09561 \times 10^{-18}}{1.6 \times 10^{-19}}$

eV

hence the kinetic energy of electron in electron volts is K = 19.35 eV.