Calculate the de Broglie Wavelength
What is the de Broglie wavelength of a 0.998 keV electron, photon, and neutron?
The de Broglie wavelength of a 0.998 keV electron is 3.86 x 10^-11 meters, the de Broglie wavelength of a 0.998 keV photon is 2.48 x 10^-10 meters, and the de Broglie wavelength of a 0.998 keV neutron is 2.20 x 10^-12 meters.
De Broglie Wavelength Calculation
(a) 0.998 keV electron:
The de Broglie wavelength of an electron can be calculated using the formula λ = h / p, where p = √(2mE), and m is the mass of the electron, and E is its energy.
Substituting the values:
λ = 6.626 x 10^-34 J.s / √(2 x 9.109 x 10^-31 kg x 0.998 x 10^3 eV x 1.602 x 10^-19 J/eV)
λ ≈ 3.86 x 10^-11 m
(b) 0.998 keV photon:
The de Broglie wavelength of a photon can be calculated using the formula λ = hc / E, where c is the speed of light and E is the energy of the photon.
Substituting the values:
λ = (6.626 x 10^-34 J.s x 3 x 10^8 m/s) / (0.998 x 10^3 eV x 1.602 x 10^-19 J/eV)
λ ≈ 2.48 x 10^-10 m
(c) 0.998 keV neutron:
The de Broglie wavelength of a neutron can be calculated using the same formula as for an electron: λ = h / p, where p = √(2mE), m is the mass of the neutron, and E is its energy.
Substituting the values:
λ = 6.626 x 10^-34 J.s / √(2 x 1.675 x 10^-27 kg x 0.998 x 10^3 eV x 1.602 x 10^-19 J/eV)
λ ≈ 2.20 x 10^-12 m
In conclusion, the de Broglie wavelength of a 0.998 keV electron is 3.86 x 10^-11 meters, the de Broglie wavelength of a 0.998 keV photon is 2.48 x 10^-10 meters, and the de Broglie wavelength of a 0.998 keV neutron is 2.20 x 10^-12 meters.