Question Video: Graphically Relating Momentum to the de Broglie Wavelength Physics

Video Transcript

The graph shows a number of curves. Which curve shows the relation between the momentum of a particle and its de Broglie wavelength?

The graph referred to in the question is this one here, with momentum on the horizontal axis and wavelength on the vertical axis. We are particularly interested in which curve is the proper graphical representation of the mathematical relationship between the momentum of a particle and its de Broglie wavelength. So, in order to figure this out, we will need to know that mathematical relationship.

That particular relationship is that 𝜆, the wavelength of the particle, is equal to ℎ, the Planck constant, divided by 𝑝, the particle’s momentum. What we can see immediately from this formula is that the wavelength of a particle and its momentum are inversely proportional. Qualitatively, this means that as momentum increases, wavelength decreases and as momentum decreases, wavelength increases. More precisely, the wavelength approaches zero with increasing momentum, and with decreasing momentum, the wavelength increases without bound.

Looking at our graph, only the red line and the yellow line show a wavelength that approaches zero with increasing momentum. So, only the red and yellow lines satisfy our first condition. But of the red and yellow lines, only the red line appears to increase without bound as momentum approaches zero. The yellow line vary clearly reaches a finite value at zero momentum. So, of the red and yellow lines, only the red line also satisfies the second condition. So, since the red Line is the only line on the graph that shows an inversely proportional relationship, it must be the curve that shows the relation between the momentum of a particle and its de Broglie wavelength.

There is another subtlety that we could have exploited to help us answer this question. And it’s a powerful idea in physics, so it’s worth exploring. If we multiply both sides of the de Broglie relationship by the momentum, we get that 𝜆 times 𝑝 equals ℎ. This is just another way to write an inversely proportional relationship. The product of our two quantities is equal to some constant. What we can notice, though, is that if we exchange 𝜆 and 𝑝, we get exactly the same formula. This means that the relationship that we have is symmetric in 𝜆 and 𝑝.

What this symmetry tells us is that whether we graph 𝜆 on the vertical axis and 𝑝 on the horizontal axis or 𝑝 on the vertical axis and 𝜆 on the horizontal axis, our graph will look the same. In turn, this tells us that whichever graph we choose to look at, the graph must be symmetric about the line 𝜆 equals 𝑝. Looking back at our graph, we see that only the blue line and the red curve have this symmetry. All the other lines would look different if we interchanged the quantities on the horizontal and vertical axes.

So, just by noticing a symmetry in our equation, we were able to greatly simplify our task, reducing our answer choices down to two. Finally, we can choose the correct answer as the red line by noting that the blue Line cannot be the correct answer because it passes through the point zero, zero. But zero times zero is not equal to the Planck constant. This alternative approach shows us that by carefully considering symmetries in an equation, we can greatly simplify problems of physical interest.