Electrons with an energy of 0.842 eV are incident on a double slit in which the two slits are separated by 40.0 nm. What is the de Broglie wavelength (in nanometers!) of these electrons?
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- Electrons with an energy of 0.610 eV are incident on a double slit in which the two slits are separated by 40.0 nm. (a) What is the speed of these electrons? m/s (b) What is the de Broglie wavelength (in nanometers!) of these electrons? nm (c) What is the angle between the two second-order maxima in the resulting interference pattern?An electric current through hydrogen gas produces several spectral lines from the transitions to n = 2, called the Balmer series. The four most visible ones are at 410.2 nm, 434.0 nm, 486.1 nm, and 656.3 nm. Assume you are observing these lines with a diffraction grating of 5870 lines per centimeter. a) What is the angle (in degrees) of the second-order maximum for the 410.2 nm line? b) What is the angle (in degress) of the second-order maximum for the 434.0 nm line? c) What is the angle (in degrees) of the second-order maximum for the 486.1 nm line? d) What is the angle (in degrees) of the second-order maximum for the 656.3 nm line?A single beam of electrons shines on a single slit of width 8.7nm. A diffraction pattern (of electrons!) is formed on a screen that is 3.9m away from the slit. The distance between the central bright spot and the first minimum is 5.7cm.What is the kinetic energy (keV, i.e. kilo electron-Volts) of the electrons?Make use of the small angle approximation.