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Theoretical limit of resolution for an electron microscope | Giancoli, Chapter 37 Ex. 43

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What's the theoretical limit of resolution for an electron microscope whose electrons are accelerated through 85kV?

The solution manual proposes the following method:

We know the following relativistic equations:

$$E^2 - (pc)^2 = m^2c^4$$

$$E = KE + mc^2$$

Thus combining them we can get $\lambda$:

$$(KE + mc^2)^2 = m^2c^4 + (pc)^2$$

$$KE^2 + 2Kmc^2 = (pc)^2 = \frac{hc}{\lambda}$$

$$\lambda = \frac{hc}{\sqrt{2KEmc^2+KE^2}}$$

Plugging numbers in we get:

$$\lambda = \frac{hc}{\sqrt{2KEmc^2+KE^2}}= 4pm$$

I agree with it.

However I approached the problem with another method.

But first we need a relation between the speed of the electron and light:

$$85000 eV\frac{1.6 \times 10^{-19} J}{1 eV} = 1/2 m_e v^2$$

From the above equation we get $v$. So:

$$v/c = 0.58$$

Now using De Broglie's wavelength with relativistic momentum:

$$\lambda = \frac{h}{p} = \frac{h\sqrt{1- (v/c)^2}}{m_e v} = 3.4pm$$

My method varies significantly from book's one. Why am I wrong?

You may argue it is wrong because I used $KE=\frac12 mv^2$ to get $\frac{v}{c}$. However I used the same method for a relativistic electron before (33kV) and got the right answer. I suspect there's a threshold (above 33kV and below 85kV).

asked Jul 24, 2019 in Physics Problems by Jorge Daniel (716 points)
edited Aug 13, 2019 by sammy gerbil

1 Answer

2 votes
Best answer

Your alternative method uses the non-relativistic approximation twice : first when you insert a value for $v/c$ in the numerator, second when you insert a value for $v$ in the denominator. Because of this your error is compounded, so you get a result which is less accurate than if you inserted the approximation only once.

Whenever you make approximations, it is most accurate to do so only once, at the last possible step, as follows :

  1. First identify a dimensionless variable $y$ which depends on the independent variable in your problem. Examples : $\beta=v/c$ if the independent variable is $v$ or $x=T/mc^2$ if it is $T=eV$( see below).

  2. Obtain a formula for the quantity you are interested in, as a function of $y$.

  3. Expand the formula into a power series in $y$, using Taylor's Theorem. Check that the expansion is valid for the range of values of $y$ which interest you. For example, the expansion $$(1+y)^n=1+ny+\frac12 n(n-1)y^2+\frac16n(n-1)(n-2)y^3+...$$ is valid provided that$y\lt 1$.

  4. Finally throw out higher powers of the dimensionless variable, depending on how much accuracy you want to have.

The accurate formula which you got can be written as $$\lambda=\frac{hc}{\sqrt{2Tmc^2+T^2}}=\lambda_0 (1+\frac12 x)^{-1/2}= \lambda_0 (1-\frac14 x+\frac{3}{32}x^2-\frac{5}{128}x^3-...)$$ where $\lambda_0=\frac{h}{\sqrt{2mT}}$ is the de Broglie wavelength assuming the electron is non-relativistic, and $x=\frac{T}{mc^2}=\frac{eV}{mc^2}$. This expansion is valid because $T=85keV$ and $mc^2=511keV$ (the rest mass of the electron) so $\frac12 x\lt 1$.

Note that the approximation is introduced only at the last step, by ignoring the higher powers of $x$. The full infinite series expansion, and the steps before it, are 100% accurate.

Using the figures given, we get $\lambda_0=4.2066pm, x=0.08317$ and $\lambda\approx4.0317pm$ when we ignore terms in $x^2$ or higher powers. Using the exact equation we get $\lambda=4.0419pm$.

Your method first makes the classical approximation $T=eV\approx \frac12 mv^2$ before any. Then $$\frac{v^2}{c^2}\approx\frac{2T}{mc^2}=4x$$ $$mv\approx\sqrt{2mT}$$ $$\lambda=\frac{h\sqrt{1-\frac{v^2}{c^2}}}{mv}\approx \lambda_0 (1-4x)^{+1/2}\approx \lambda_0 (1-2x-2x^2+4x^3-...)$$

Compare this power series with the one above. The term in $x$ now has a coefficient of$2$ instead of $\frac14$, so the correction to $\lambda_0$ is $8\times$ what it should be. The previous line was already an approximation, so the power series is also an approximation, regardless of however many terms we retain. Unlike the series above, which gets close to being 100% accurate as we retain more and more powers.

Ignoring terms in $x^2$ or higher powers. we get $\lambda \approx 3.5069pm$ instead of$\lambda \approx 4.0317pm$.

The larger is the value of $T=eV$ then the larger is $x$ also, and the greater is the difference between the correct approximation $\lambda\approx \lambda_0 (1-\frac14 x)$ and your incorrect approximation $\lambda\approx \lambda_0 (1-2x)$. And as noted above, if $x\gt \frac14$ then your approximation $\lambda\approx \lambda_0 (1-4x)^{+1/2}$ gives an imaginary result, whereas the exact formula $\lambda= \lambda_0 (1+\frac12x)^{-1/2}$ gives a real result because $x\ge 0$ for all values of $V$.

There is no sharp cut off point for $V$ at which the electron suddenly switches over from being classical to relativistic, or above which your approximation suddenly ceases to be a good one. The change is gradual.

At $V=33kV$ the value of $x$ is less than half that at $V=85kV$ so the smaller is the difference between your erroneous calculation and the correct one. At both values of $V$ your method gives an error which is $8\times$ bigger than it should be if you used the correct approximate formula.

In fact, if the accelerating potential is such that $T=eV \gt \frac12 mc^2$ then $v \gt c$ so the factor $\sqrt{1-\frac{v^2}{c^2}}$ in the numerator would be imaginary, and so would the de Broglie wavelength.

answered Aug 13, 2019 by sammy gerbil (28,896 points)
selected Aug 14, 2019 by Jorge Daniel
Thanks for this really helpful answer. You refer to figures in your post but I do not see them.
I mean numbers  - ie the value(s) of the potential $V$.
I get $x=0.16634$ instead of $x = 0.08317$ (knowing that $x=\frac{T}{mc^2}=\frac{eV}{mc^2}$). I may be mistaken but didn't you mean $y=0.08317$ Instead of $x = 0.08317$?