A parallel beam of electrons accelerated by a potential difference of 25 kV falls along a normal on an aperture with two slits. The distance between the slits is 50 µm. Find the distance between maximums on the screen located at the distance of 1 m from the slits.
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- Consider scattering with electrons of 54 eV on a crystal with planes separated by 0.22 nm. How many diffraction peaks should be expected? Justify your answerA highly collimated (parallel) beam of electrons is shot through a single slit of width 17.5μm. The electrons are moving with a speed of 7.357.35km/s. When they hit the screen, located at distance 1.57m away, the distribution of hitting positions makes a pattern with a central peak and minima on either side. What is the width of the central peak (equivalently, distance between the minima on either side)? The mass of an electron is 9.11×10^−31 kg.A beam of electrons with kinetic energy 45 keV is shotthrough two narrow slits in a barrier. The slits are a distance 2.0 x 10-6 m apart. If a screen is placed 45.0 cm behind the barrier, calculate the spacing between the “bright” fringes of the interference pattern produced on the screen
- A beam of electrons strikes a double slit, with the slit separation of 102 nm. The first-order diffraction maximum is seen at an angle of 5 degrees away from the beam's direction. What was the speed of the electrons? (Hint: First determine the wavelength of the electrons using the double-slit interference equations from the optics unit. Then use that to determine the speed of the electron.)Please explain in detail with steps how to set up and solve for this equation. Find the de Broglie wavelength for a 1500kg car when its speed is 80km/h. How significant are the wave properties of this car likely to be?Molybdenum has a work function of 4.20 eV. What is the stopping potential if the incident light has a wavelength of 180 nm?
- For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is (a) a photon; and (b) an electron.In the classical limit calculate the wavelength corresponding to an electron with the energy of 99 kev (kiloelectronvolt). Give your answer in Angstrom (10-10 1.6x10-10 m, then write 1.6 as your answer). This m, for example, if the answer is should give you a good idea why one can use a crystal lattice with an average interatomic distance of around 1010 m to observe electron diffraction.I need help with this question. Originally I got 3 degrees for the answer, but it appears that this is incorrect and I don't know what went wrong. Here is the question: Electrons with an energy of 0.610 eV are incident on a double slit in which the two slits are separated by 60.0 nm. Electron speed is 4.63e+05 m/s and the de Broglis wavelength of the electrons is 1.57 nm. What is the angle between the two second-order maxima in the resulting interference pattern. I really appreciate the help!