Here is the problem:

Find the energy of neutron, electron and electromagnetic waves of wavelength 0.1nm.

English is my second language, so I am kind of confused about the meaning of problem itself. So, to find the energy of neutron, and electron, just use Einstein's energy equation $E= mc^2$? Since it doesn't point out kinetic energy, I will just assume that neutron and electron are rest, and don't have to consider relativity like $E=\sqrt{p^2c^2+m^2c^2}$? For electromagnetic waves, I will just use the equation $E=hc/\lambda$?


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Here is a hint - de Broglie wavelength for the neutron and electron - does that help?

  • $\begingroup$ How can I get the waavelength for the neutron without knowing its energy or speed something? $\endgroup$ – kengkeng Nov 12 '14 at 21:45
  • $\begingroup$ @kengkeng This is a good starter. Check out $\lambda = \frac{h}{p}$ and $f = \frac{E}{h}$. $\endgroup$ – HDE 226868 Nov 12 '14 at 21:50
  • $\begingroup$ @kengkeng you are told the wavelength is 0.1nm - so you can use de Broglie wavelength equation....- first equation shown in comment of HDE226868 - h is a constant and p is momentum - you should be then able to work from momentum to energy.... does this help? $\endgroup$ – tom Nov 12 '14 at 22:21
  • $\begingroup$ @tom yeah! thanks! The my main concern of this question is grammar... I thought 0.1nm was only for em waves. thanks! $\endgroup$ – kengkeng Nov 12 '14 at 22:23

The idea behind this exercise is, to show you that everything CAN act like a wave in quantum mechanics, see slit experiments. You will have to use the same formula for all three entities. As stated before, the de Broglie wavelength is the way to go when confronted with such tasks.

For the non-relativistic case, the de Broglie formula states:

$$ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2*m*E_{kin}}} $$ I'm letting you solve the equation as an exercise.

Source: Master Physics Student

  • $\begingroup$ I know it sounds stupid, but when you have "of wavelength 0.1nm", it means all electron, neutron, and em waves have this wavelength, right? $\endgroup$ – kengkeng Nov 12 '14 at 22:21
  • $\begingroup$ Yes. The Energie will differ because of their masses. For the photon of course you already stated the correct formula in your question. $\endgroup$ – qacwnfq q Nov 12 '14 at 22:23

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