Question

Determine the wavelength of an electron that has a kinetic energy of 1.15 × 10^{-19} J. Report your answer in units of nm to the correct number of significant figures.

Answer 1

Answer:

The electron has a wavelength of 1448 nanometers.

Explanation:

The wavelength of the electron can be determined by means of the de Broglie wavelength.

$\lambda = \frac{h}{p}$  (1)

Where h is the Planck's constant and p is the momentum

$\lambda = \frac{h}{mv}$ (2)

The velocity can be found by means of the kinetic energy equation.

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Therefore, equation 3 can be replaced in equation 2

$\lambda = \frac{h}{\sqrt{2mKE}}$ (4)

Where m is the mass of the electron and KE its kinetic energy.

$\lambda = \frac{6.624x10^{-34} J.s}{\sqrt{2(9.1x10^{-31}Kg)(1.15x10^{-19} J)}}$

But $1J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34} Kg.m^{2}/s^{2}.s}{\sqrt{2.093x10^{-49}Kg^{2}.m^{2}/s^{2}}}$

$\lambda = \frac{6.624x10^{-34} Kg.m^{2}/s^{2}.s}{4.574x10^{-25}Kg.m/s}$

$\lambda = 1.448x10^{-9}m$

Finally,  the wavelength can be expressed in terms of nanometers:

$\lambda = 1.448x10^{-9}m .\frac{1x10^{9}nm}{1m}$ ⇒ $1448nm$

Hence, the electron has a wavelength of 1448 nanometers.