A 0.22-caliber handgun fires a 1.9-g bullet at a velocity of 765 m/s. Calculate the de Broglie wavelength of the bullet. Is the wave nature of matter significant for bullets?

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hey everyone in this example, we're working with a helium atom moving at the following speed and we need to find the d broccoli wavelength in PICO meters for our helium atom. So what we should recognize is that based on the speed they gave us, this is actually going to be our velocity here. And we should recognize that it's in the proper units as meters per second. And also because we need to find wavelength. To answer this question, we want to recall our formula for wavelength, which is wave length is equal to plank's constant divided by the mass of our helium atom, multiplied by the helium atoms given velocity. So what we want to reference is our periodic tables to find helium. And we'll see helium in group eight A where we're going to make note of the molar mass Which is listed as 4. g per mole. And so when it comes to the mass for our calculation for wavelength Broccoli Wavelength, we want to go ahead and actually convert this two kg Permal. So that's going to be our first step. We're going to take the Periodic table Molar Mass 4. grams Permal of our helium. And we're going to first go from grams to kilograms. So we want to recall that for one kg. Our prefix tells us we have 10 to the third power grams. This allows us to cancel out grams and now we're left with kilograms Permal, which is what we want for our mass when it comes to finding the Wavelength. So this calculation gives us a value of 4.03 times 10 to the negative third power kg per mole as our proper mass for our helium atom. And so now that we have our proper mass units, we're going to go ahead and calculate the mass of a helium atom. And so what we're going to do is take our molar mass in kilograms per mole four point oh three times 10 to the negative third power kilograms Permal of our helium. And we're going to want to go ahead and recall avocados number as a conversion factor so that we have for one mole 6.02, 2 times 10 to the 23rd power helium atoms. And so this allows us to get the mass of one helium atom so we can go ahead and cancel out our units of moles. And this leaves us with just our mass for a helium atom in kilograms. And what we're going to get is a result of 6.6473 times to the negative 27th power kilograms. So now that we have the mass of our helium atom, we're going to go ahead and go into our calculation for wavelength. So we'll have that wavelength is equal to in our numerator plank's constant, which we should recall is 6.6 26 times 10 to the negative 34th power jewels time seconds. And then in our denominator we have our mass of our helium atom that we just calculated 6.6473 times 10 to the -27th power kg. And then this is multiplied by our velocity. And the problem which is given to us as 1,537 m/s. And so our next step is to recall that a jewel is also interpreted as kilograms times meters squared divided by second squared. So we're going to actually substitute this in place of our jewels term to simplify our numerator. And so what we would have is that our wavelength is equal to six 626 times 10 to the negative 34th power substituting in the unit for jules, we should have kilograms times meters squared divided by seconds squared. And then we're still being multiplied by seconds as we were in the numerator here. And then we're still going to be dividing by our denominator, which we can actually just simplify by multiplying these two quantities and that's going to give us a value equal to 1.23 times 10 to the negative 23rd power kilograms times meters divided by seconds. And so now our next step is to simplify our units. So we can go ahead and cancel out kilograms because we have it in the numerator and the denominator. We can also cancel out seconds in our numerator As well as in our denominators to fully cancel out seconds. And then we can cancel out meters leaving us with one m to the first power in our numerator. And so our last step is to calculate this quantity here in our calculators. And this is going to give us our wavelength equal to a value of 6. times 10 to the negative 11th power in units of meters. However, according to the problem, we want our final answer to be in PICO meters. So we're going to recall the conversion factor to go from meters to PICO meters. And so our prefix PICO tells us that we should have 10 to the negative 12 power of our base unit meters. So this allows us to cancel out meters leaving us with our final unit pick zero m for our wavelength. And this is going to give us our final answer equal to 64.8 PICO meters as our wavelength for our helium atom. And so this here completes our final answer for this question. So if you have any any other questions as to how we solve this problem, just leave them down below. And I will see everyone in the next practice video

A 0.22-caliber handgun fires a 1.9-g bullet at a velocity of 765 m/s. Calculate the de Broglie wavelength of the bullet. Is the wave nature of matter significant for bullets?

Verified Solution

Key Concepts

de Broglie Wavelength

Momentum

Wave-Particle Duality