Question Video: Identifying the de Broglie Relationship Physics • 9th Grade

Which of the following formulas shows the relation between the de Broglie wavelength of a particle, 𝜆, its momentum 𝑝, and the Planck constant ℎ? [A] 𝜆 = 𝑝/ℎ [B] 𝜆 = ℎ𝑝² [C] 𝜆 = ℎ/𝑝² [D] 𝜆 = ℎ/𝑝 [E] 𝜆 = ℎ²𝑝²

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Video Transcript

Which of the following formulas shows the relation between the de Broglie wavelength of a particle 𝜆, its momentum 𝑝, and the Planck constant ℎ? (a) 𝜆 equals 𝑝 divided by ℎ. (b) 𝜆 equals ℎ times 𝑝 squared. (c) 𝜆 equals ℎ divided by 𝑝 squared. (d) 𝜆 equals ℎ divided by 𝑝. (e) 𝜆 equals ℎ squared times 𝑝 squared.

The question is asking us to identify the formula for the de Broglie wavelength of a particle. And we are told what our variables mean. The wavelength of the particle is 𝜆, the momentum is 𝑝, and the Planck constant is ℎ. The easiest way to answer this question correctly is to know the de Broglie formula by heart. And if we know this, we can immediately identify (d) as the correct answer.

But if we don’t know this formula by heart, we can still figure out that (d) is the correct answer by paying attention to the dimensions of our quantities. 𝜆 is a wavelength, which unsurprisingly has dimensions of length. We’ll use the capital letter L to represent length. And what it means for the dimensions of 𝜆 to be length is that if we reported a value for 𝜆, we would use units like meters or centimeters or kilometers. But we would never use units like seconds or columns or amperes.

Similarly, momentum is mass times velocity, and velocity is distance per time. So the dimensions of momentum are mass times length divided by time. The capital letter M means mass, and the capital letter T means time. And just as a matter of convention, we write T to the negative one power rather than one divided by T.

Finally, the Planck constant, we may recall that one common value for the Planck constant is 6.626 joules seconds. Joules are a unit of energy, and seconds are a unit of time. So the dimensions of the Planck constant are energy times time. To express energy in terms of the dimensions that we have already used, length, mass, and time, we can recall the formula for kinetic energy: one-half mass times velocity squared. As we can see from this formula, the dimensions of energy are the same as mass times the dimensions of velocity squared. And we know the dimensions of velocity are length per time. So the dimensions of energy are mass times length squared per time squared. These dimensions are not just for kinetic energy, but for any energy. So the Planck constant with units of energy times time has dimensions of mass times length squared times time to the negative one power.

Now that we know the dimensions of each of the three quantities that we’re working with, we are ready to work out the answer to the question. First, we observe that mass and time do not appear in the dimensional formula for the wavelength. But they do appear in the dimensional formulas for momentum and the Planck constant. So the final formula that we’re looking for must combine ℎ and 𝑝 in such a way that the mass from momentum cancels the mass from the Planck constant and time from momentum cancels the time from the Planck constant, leaving overall dimensions of length.

Looking at our available formulas then, we can immediately eliminate choice (b) ℎ times 𝑝 squared and choice (e) ℎ squared times 𝑝 squared as possible answers. This is because we need to eliminate mass and time from our final dimensional formula. But any time we multiply 𝑝 by ℎ, we wind up with more factors of mass and time because mass times mass is mass squared and time to the negative one times time to the negative one is time to the negative two. So our final formula needs to include division, not multiplication.

We can use a similar argument to eliminate the choice ℎ divided by 𝑝 squared. The dimensions of 𝑝 squared are M squared times L squared times T to the negative two, just doubling all the exponents in the dimensional formula for 𝑝. But as we can see, 𝑝 squared actually has one more factor of mass and time than ℎ does. So dividing ℎ by 𝑝 squared still leaves us with overall dimensions that include mass and time. This leaves us with 𝑝 divided by ℎ and ℎ divided by 𝑝. Since these two are reciprocals of one another, once we find the dimensions of one of them, we’ll know the dimensions of the other by just changing the sign on all of the exponents.

So let’s find the dimensions of 𝑝 over ℎ. The dimensions of 𝑝 over ℎ are given by M times L times T to the negative one times M to the negative one times L to the negative two times T. The first three dimensions are from 𝑝, and the second three are from ℎ, with the sign reversed on each exponent because dividing by something is the same as multiplying by that thing raised to the negative one power. When we simplify this formula, we note that M times M to the negative one power is dimensionless, as is T to the negative one power times T. L times L to the negative two is equal to L to the negative two plus one, which is just L to the negative one. So the dimensions of 𝑝 over ℎ are L to the negative one power.

Now, this is not what we need because we’re looking for dimensions of L. However, it is very close. It’s the reciprocal of what we need. This means that ℎ over 𝑝, the reciprocal of 𝑝 over ℎ, will have dimensions of L, which is exactly what we’re looking for. Although, as we’ve shown, we can derive the correct answer using dimensional analysis, the de Broglie wavelength relation is so important and so simple that it is worth just memorizing and knowing the answer without having to do any work.