I have a problem that asks for the minimum energy of a wave that we will use to see a particle of size $.1\text{ nm}$. I understand that I can not see a $.1\text{ nm}$ particle with any wave length larger than $.1\text{ nm}$. I thought this would be easy, and I would use De Broglie's relation of electron waves,

$$f=\frac{E}{h}\quad\text{or}\quad E=fh=\frac{hc}{λ}$$

Using this I get $12400\text{ eV}$... this is the wrong answer.

What the book says to do is use an equation "wavelength associated with a particle of mass $M$." It is:


OR for my specific case:

$$λ=\frac{1.226}{\sqrt{K}}\text{ nm}$$

This second equation, if I'm correct, is getting the kinetic energy of the wavelength, not the total energy.

I do not understand what I should be looking for in problems asking for energy of wavelengths to distinguish the use of the first equation I presented vs. the second one. Any enlightenment on this area would be appreciated.


2 Answers 2


Good question, actually. The way you thought to do the problem at first is fine conceptually, but you used the wrong equation: $E=\frac{hc}{\lambda}$ applies only to photons or other massless particles. The actual equation for computing De Broglie wavelength is

$$\lambda = \frac{h}{p}$$

where $p$ is momentum. You can then put that in terms of energy using $E = pc$ for a photon (which will give you $\lambda = \frac{hc}{E}$), or $E = mc^2 + \frac{p^2}{2m}$ for a nonrelativistic particle (like a slow-moving electron), or $E = \sqrt{m^2c^4 + p^2c^2}$ for anything in general.

If you're curious, you can derive the equation the book uses from $\lambda = \frac{h}{p}$ by using the expression for a nonrelativistic particle. Just remember that $K$ is kinetic energy, so $E = mc^2 + K$. Also, you'll have to drop a relatively small term.


You can definitely image an object that is much smaller than the wavelength you use to image it!

Using a regular microscope, one can't indeed do much better, as the Abbe limit states. I think that's the limit you were implicitly referring to when you said the wavelength had to be larger than $0.1 nm$. See


However, there are many other ways to circumvent this limit, which is more technical than truly fundamental. See the long article on sub-diffraction microscopy on Wikipedia: http://en.wikipedia.org/wiki/Super-Resolution_microscopy


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