please explain, thank you! The electron beam in a certain electron microscope produces electrons with a...
The electron beam in a typical scanning electron microscope produces kinetic energies of 27.0 keV (a) Assuming that relativistic effects can be ignored, what is the speed of the electrons? Number m/s (b) What potential difference is required to produce electrons of this energy? Number Units (c) This accelerating potential difference is applied over a distance of 1.80 cm. What is the electric field in that region? Number Units
Electrons in an electron microscope have a kinetic energy of 6.01 105 eV. (a) Find the de Broglie wavelength of the electrons. (b) Find the ratio of this wavelength to the wavelength of light at the middle of the visible spectrum (550 nm). (c) How many times greater magnification is theoretically possible with this microscope than with a light microscope?
The electron beam in a typical scanning electron microscope produces kinetic energies of 22.0 keV. (a) Assuming that relativistic effects can be ignored, what is the speed of the electrons? 8.791 10m/s (b) What potential difference is required to produce electrons of this energy? Number Units (c) This accelerating potential difference is applied over a distance of 2.00 cm. What is the electric field in that region? NumberUnits
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")?...
Consider electrons of kinetic energy 5.29 eV and 529 keV. For each electron, find the de Broglie wavelength (in nm), particle speed (in m/s), phase velocity (speed, in m/s), and group velocity (speed, in m/s). nm 5.29 eV electron de Broglie wavelength particle speed phase velocity group velocity m/s m/s m/s 529 keV electron nm de Broglie wavelength particle speed phase velocity group velocity m/s m/s m/s
(10 pts) An electron beam, for cancer therapy, was configured by accelerating the electrons to 1.00 x 105 eV kinetic energy. What is the de Broglie wavelength of these electrons, in meters, to three sig. figs.? 1/m =
11. (10 pts) An electron beam, for cancer therapy, was configured by accelerating the electrons to 1.00 x 105 eV kinetic energy. What is the de Broglie wavelength of these electrons, in meters, to three sig. figs.? W/m =
11. (10 pts) An electron beam, for cancer therapy, was configured by accelerating the electrons to 1.00 x 105 eV kinetic energy. What is the de Broglie wavelength of these electrons, in meters, to three sig. figs.? W/m =
An electron beam was configured by accelerating the electrons to 1.00 × 10^5 eV kinetic energy. What is the de Broglie wavelength of these electrons, in meters?
11. (10 pts) An electron beam, for cancer therapy, was configured by accelerating the electrons to 1.00 x 105 eV kinetic energy. What is the de Broglie wavelength of these electrons, in meters, to three sig. figs.?