K = m c^2 - m0 c^2
and E = mc^2 = sqrt [ p^2 c^2 + m0^2 c^4]
and mc^2 = k + m0c^2
putting in,
k + m0c^2 = sqrt [ p^2 c^2 + m0^2 c^4]
square both sides,
k^2 + m0^2 c^4 + 2km0c^2 = p^2 c^2 + m0^2 c^4
k^2 + 2k m0c^2 = p^2 c^2
p = sqrt[ k (k + 2m0 c^2) ] / c
lambda = h / p
lambda = h / (sqrt[ k (k + 2m0 c^2) ] / c )
lambda = hc / sqrt[ k (k + 2m0 c^2) ]
For relativistic particles the de Broglie relation lambda_dB = h/p still holds, but recall that the...
De Broglie postulated that the relationship ? = h/p is valid for relativistic particles. What is the de Broglie wavelength for a (relativistic) electron having a kinetic energy of 3.39 MeV? answer in m
De Broglie postulated that the relationship λ = h/p is valid for relativistic particles. What is the de Broglie wavelength for a (relativistic) electron having a kinetic energy of 3.49 MeV?
The de Broglie wavelength calculation for an object holds even at relativistic speeds. At what total energy E will the de Broglie wavelength of an electron be different by a factor of two from the wavelength of a photon with the same energy?
(a) Using the relativistic relation between E and p, show that electrons and photons with the same energy E have different wavelengths. (Note: Even at relativistic energies, the de Broglie wavelength equation is valid.) (b) Show that the ratio of their wavelengths approach equality as their common energy E gets much larger than mec^.
a) Discuss why the de Broglie wavelength λ corresponding to a momentum p (p wavenumber given by k # 2n/A) leads to a representation of p by the operator p as (h/) (d/dx) hk, where k is the b) Using theoperao orm of p given in part a, show that,pih c) The total energy of a simple harmonic oscillator of mass M and spring constant K can be written as H- p2/M + ke . If the mass is displaced...
Learning Goal: To understand de Broglie waves and the calculation of wave properties. In 1924, Louis de Broglie postulated that particles such as electrons and protons might exhibit wavelike properties. His thinking was guided by the notion that light has both wave and particle characteristics, so he postulated that particles such as electrons and protons would obey the same wavelength-momentum relation as that obeyed by light: λ=h/p, where λ is the wavelength, p the momentum, and h Planck's constant. Part...
Find the de Broglie wavelength λ for an electron moving at a speed of 1.00×106m/s. (Note that this speed is low enough that the classical momentum formula p=mv is still valid.) Recall that the mass of an electron is me=9.11×10−31kg, and Planck's constant is h=6.626×10−34J⋅s.
Part B The hypothesis that was put forward by Louis de Broglie in 1924 was astonishing for a number of reasons. An obvious reason is that associating a wavelike nature with particles is far from intuitive, but another astonishing aspect was how well the hypothesis fit in with certain parts of existing physics. In this problem, we explore the correspondence between the de Broglie picture of the wave nature of electrons and the Bohr model of the hydrogen atom. What...
What is the de Broglie wavelength (in meters) of a neutron traveling at a speed of 0.92 c? Since the neutron's speed is close to the speed of light (c), Special Relativity must be used when calculating the linear momentum (p). The mass of the neutron is 1.675 x 10-27 kg. Suppose that an alpha particle (mαα = 6.646 x 10-27 kg) has a kinetic energy of 75 keV. What is the alpha particle's speed (v) (in terms of "c")?...
Derive the following relation: h/mec (1 − cos θ) = λ ′ − λ It is suggested you use the following strategy: First, use the momentum equations and the relation cos2 φ + sin2 φ = 1 to eliminate φ. Next use the energy equation and the relativistic relation E^2 = m^2 c^4 + p^2 c^2 to find an expression for the square of the momentum of the electron that does not depend on v (or γ). Finally, use this...