Question

An electron microscope produces electrons with a 2.00-pm wavelength. If these are passed through a 1.00-nm single slit, at what angle will the first diffraction minimum be found?

$ 0.115^\circ $

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This is College Physics Answers with Shaun Dychko. An electron microscope produces electrons that have a de Broglie wavelength of 2 picometers. Now these electrons go through a slit with a size of 1 nanometer. And the question is, where's the first diffraction minimum? What is the angle to that diffraction minimum? So here's the slit that the electrons are going through and it's going to make a central maximum and where this first minimum occur, at this angle here. This is back in chapter 27 that we studied this and the equation 21 is what we are going to use and that says the slit width times sin of this angle equals some integer times the wavelength. This isn't strictly an integer because zero is not included. So anyway. We want to choose the number 1 for the letter

*M*here. Then also, we'll solve for*sin Θ*by dividing both sides by*D*. So*sin Θ*is*Mλ*over*D*, in which case, Θ is the inverse sin of both sides; inverse sin of*Mλ*over*D*and then you take the inverse sin of 1 because this is the first minimum that we are looking for times the wavelength and pico is times 10 to the minus 12 and you divide that by the slit width of 1 times 10 to the minus 9 meters and we get an angle of 0.115 degrees.