What is the formula for electron speed from kinetic energy?+
The relativistic formula is: first compute the Lorentz factor gamma = 1 + K/(m_e*c^2) = 1 + K(eV)/511000, then v = c * sqrt(1 - 1/gamma^2). Here m_e*c^2 = 511 keV is the electron rest energy. For K = 50 keV: gamma = 1.0979, v = c * sqrt(1 - 0.8294) = 0.4131c = 1.24 x 10^8 m/s. The classical formula v = sqrt(2K/m_e) is accurate only below about 5 keV.
What is the electron rest energy and how does it affect calculations?+
The electron rest energy m_e*c^2 = 511 keV = 0.511 MeV. This is the reference scale for relativistic effects. When kinetic energy K equals the rest energy (511 keV), gamma = 2 and v = 0.866c (86.6% of light speed). When K is much smaller than 511 keV, relativistic effects are negligible and the classical formula applies. When K is comparable to or larger than 511 keV, the full relativistic treatment is required.
What is the Lorentz factor and what does gamma mean?+
The Lorentz factor gamma = 1/sqrt(1 - v^2/c^2) = 1 + K/(m_e*c^2) quantifies relativistic effects. At rest, gamma = 1. At v = 0.5c, gamma = 1.155. At v = 0.866c, gamma = 2. At v = 0.9950c, gamma = 10. Physically, gamma is the ratio of total energy to rest energy (E = gamma*m_e*c^2), and it appears in time dilation (t' = gamma*t), length contraction (L' = L/gamma), and relativistic momentum (p = gamma*m_e*v).
What is the de Broglie wavelength of an electron at different energies?+
The relativistic de Broglie wavelength lambda = hc/sqrt(K^2 + 2*K*m_e*c^2), where K is kinetic energy. Representative values: 1 eV electron lambda = 1.23 nm (infrared), 100 eV lambda = 0.123 nm (X-ray range), 1 keV lambda = 38.8 pm, 50 keV lambda = 5.36 pm, 200 keV lambda = 2.51 pm (electron microscope), 1 MeV lambda = 0.872 pm, 10 MeV lambda = 0.124 pm. The small wavelength of keV to MeV electrons enables atomic-resolution imaging.
When must I use the relativistic formula instead of the classical one?+
Use the relativistic formula whenever the kinetic energy K exceeds about 5 keV (1% of the rest energy). At K = 5 keV, the classical formula overestimates speed by about 1%. At K = 50 keV, the overestimate is 7%. At K = 511 keV (equal to rest energy), the classical formula gives 1.41c, which exceeds the speed of light and is physically impossible. For any calculation involving X-ray tubes (40+ kV), electron microscopes (60-300 kV), or particle beams (any MeV energy), always use the relativistic formula.
What is beta in the context of electron speed?+
Beta (beta) = v/c is the dimensionless speed of the electron as a fraction of the speed of light. At rest, beta = 0. For a 50 keV electron, beta = 0.4131 (41.31% of c). At the electron rest energy (511 keV), beta = 0.8660 (86.60%). Beta approaches 1 asymptotically as kinetic energy increases, but never reaches 1. The combination beta*gamma appears in relativistic momentum: p = m_e*beta*gamma*c.
How does an electron microscope use electron speed and wavelength?+
Electron microscopes accelerate electrons to 60 to 300 keV. At 200 keV, beta = 0.695 (69.5% of c), gamma = 1.391, and the de Broglie wavelength is 2.51 pm. This is about 40 times shorter than the wavelength of X-rays used in X-ray crystallography (typically 100 pm). The shorter wavelength means the theoretical resolution limit of the electron microscope is at the sub-angstrom scale, enabling direct imaging of individual atoms and even chemical bonds in favourable cases.
What voltage accelerates an electron to 50% of the speed of light?+
At beta = 0.5 (50% of c), gamma = 1/sqrt(1 - 0.25) = 1.1547. The kinetic energy K = (1.1547 - 1) x 511 keV = 79.1 keV. Since kinetic energy in eV equals the accelerating voltage in volts, a voltage of 79.1 kV (79,100 volts) is needed. This is higher than a typical cathode ray tube voltage (15-30 kV) but within the range of older X-ray imaging systems and some electron diffraction instruments.
What is total relativistic energy versus kinetic energy?+
Total relativistic energy E = gamma*m_e*c^2 = kinetic energy K + rest energy m_e*c^2 = K + 511 keV. For a stationary electron, E = 511 keV (pure rest energy). For a 50 keV electron, E = 561 keV. For a 1 MeV electron, E = 1511 keV. The formula E^2 = (pc)^2 + (m_e*c^2)^2 is the relativistic energy-momentum relation that replaces E = p^2/(2m) from classical mechanics.
How fast are electrons in atoms and metals?+
In a hydrogen atom ground state, the electron orbits at approximately 2.19 x 10^6 m/s (0.73% of c), corresponding to about 13.6 eV kinetic energy. This is non-relativistic. Conduction electrons in metals drift slowly under an applied field (millimetres per second) but have a high Fermi velocity of about 1 to 2 x 10^6 m/s due to quantum pressure, corresponding to 3 to 10 eV. Even the fastest conduction electrons in metals are essentially non-relativistic.
Can the classical and relativistic formulas ever give the same result?+
Yes, in the limit of very low kinetic energy (K much less than 511 keV), both formulas converge to the same result. Mathematically, expanding the relativistic formula to first order in K/(m_e*c^2) gives v = sqrt(2K/m_e) plus a small correction term, which equals the classical formula. Below 1 keV, the two formulas agree to within 0.1%. The classical formula is simply the low-energy limit of the more general relativistic formula.
What is relativistic momentum and how does it differ from classical momentum?+
Classical momentum is p = m_e*v. Relativistic momentum is p = gamma*m_e*v. The two agree at low speeds (gamma near 1) and diverge increasingly at high speeds. For a 50 keV electron, p_rel = 1.0979 x p_classical (about 10% larger). For a 511 keV electron (gamma = 2), p_rel = twice the classical value. Relativistic momentum always appears in the de Broglie wavelength: lambda = h/p = h/(gamma*m_e*v). Using classical momentum overestimates the wavelength by a factor of gamma.