De Broglie Wavelength Example 1

Jules Bruno
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So here we have to find the wavelength of a proton with the speed of 6. times 10 to the 5 m per second. Here, the mass of a proton is recorded as 1.67 times, 10 to the negative 27 kg. Alright, so it's going to be wavelength equals plank's constant divided by mass times velocity. Now remember here that plank's constant is Jules Times seconds and what we need to realize here is that Jewell one Jewell is equal to kilograms times meter squared over second squared. So if we're doing Jules time seconds, that really means kilograms times meter squared over second squared times seconds. So here, one of these seconds will cancel out with this second. So the units for plants constant would be 6.626 times 10 to the negative, 34 kilograms times meter squared over seconds. We do this to see how the kilograms of my subatomic particle in this case can cancel out. So we have 1.67 times 10 to the negative 27 kg and it's traveling at a velocity of 6.25 times 10 to the 5 m per second. So here what cancels out our meters, cancels out with meters and then we're gonna have here seconds. Cancel out with seconds kilograms, cancel out with kilograms. So we're gonna have 1 m left. So when we plugged that into our calculator will get 6. times 10 to the negative 13 m for our wavelength off our proton within this question
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