Question

Calculate the de Broglie wavelength of (a) a 0.715 keV electron (mass = 9.109 × 10^-31 kg), (b) a 0.715 keV photon, and (c) a 0.715 keV neutron (mass = 1.675 × 10^-27 kg).

Answer 1

Calculate the de Broglie wavelength of (a) a 0.715 keV electron (mass = 9.109 × 10-31 kg), (b) a 0.715 keV photon, and

(b) Wavelength of a 0.715 keV photon

E = h v = h c / λ
where

E = energy of a photon (J)
h = Planck's constant = 6.626 x 10^ -34 J s
c = speed of light = 3 x 10^8 m / s
λ = wavelength (m)

(1 eV = 1 electron volt = 1.602 × 10-19 J)

E = h c / λ

[715 eV x (1.602 × 10^-19 J/eV)] = (6.626 x 10^ -34 J s)(3 x 10^8 m / s) / λ

5.762×10^8 m^(-1) = 1 / λ

λ = 1.736 x 10^-9 m

(a) Wavelength of a 0.715 keV electron

To determine the de Broglie wavelength, the following equation will be used:

λ = h / momentum = h / p = h / mv = h / √(2m x Kinetic Energy)

where

h = Planck's constant = 6.626 x 10^ -34 J s
mass (electron) = 9.109 x 10^-31 kg
λ = wavelength (m)

λ = h / √(2m x Kinetic Energy)

λ = (6.626 x 10^ -34 J s) / √{2(9.109 x 10^-31 kg) x [715 eV x (1.602 × 10^-19 J/eV)]}

λ = (6.626 x 10^ -34 J s) / √(2.087 x 10^-46)

λ = (6.626 x 10^ -34 J s) / 1.44 x 10^-23

λ = 4.6 x 10^-11 m