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:
$$λ=\frac{hc}{\sqrt{2mc^2K}}$$
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.
Best Answer
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.