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A problem (Problem 1.4) in Zettili's Intro to Quantum Mechanics is:

Estimate the energy of the electrons that we need to use in an electron microscope to resolve a separation of 0.27 nm.

I was not able to find any reference in Zettili regarding the nature of electron microscopes (prior to this exercise) so I assume that this question is based on some criteria involving the interference or diffraction of electrons. In the provided solution for the problem (in the book) the energy is simply calculated by using de Broglie's relation with a wavelength of 0.27nm as such:

\begin{equation} E = \frac{p^2}{2m_e} = \frac{2\pi^2\hbar^2}{m_e\lambda^2} \end{equation}

After googling for a bit I found that electron microscopes have a resolution on the order of the de Broglie wavelength of the electrons used, why is this so?

Thanks.

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  • $\begingroup$ Heuristically this is determined by Heisenberg's uncertainty principle $$\Delta k\Delta x \sim 1$$ where $k = 2\pi/\lambda$. Hence the precision $\Delta x$ is estimated by the de Broglie wavelength $\lambda$. $\endgroup$ – Phoenix87 Dec 27 '14 at 23:18
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    $\begingroup$ The resolution of electron microscopes is usually quite a lot worse than the de Broglie wavelength suggests. You also have to use Abbe's formula, which contains the sine of the half angle of the beam, just like in optical microscopy. Since this numerical aperture is typically small for electron optics, the resolution is a couple of orders of magnitude lower than the theoretical diffraction limit. A Zeiss 1455 is specified at 3.5nm resolution at 30keV. The de Broglie wavelength at that energy is 0.007nm, or 500 times smaller. $\endgroup$ – CuriousOne Dec 27 '14 at 23:30
  • $\begingroup$ @Phoenix87 It seems like in order to make your heuristic argument you need to assume that: \begin{equation}\Delta x\Delta p \sim \hbar \end{equation} why is this valid? I understand that the uncertainty principle calls for at least \begin{equation}\Delta x\Delta p \ge \hbar/2\end{equation} but why the jump to \begin{equation}\Delta x\Delta p \sim \hbar \end{equation} ? $\endgroup$ – Loonuh Dec 28 '14 at 4:04
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    $\begingroup$ $\sim\hbar$ from the fact that you want to "squeeze" uncertainty as much as possible, so practically you end up with something of the order $\hbar$ $\endgroup$ – Phoenix87 Dec 28 '14 at 9:46

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