What accelerating voltage would be required for the electrons of an electron microscope if the microscope is to have the same resolving power as could be obtained using 100 keV gamma rays?

Question

ganeshie8 · Answer

Assuming the minimum resolvable feature size is same as the `wavelength` of the light used in the microscope, I think this problem is about finding the required potential difference that builds up the corresponding wavelength for the electron...

kainui · Answer

Yeah I think you're right about that. I'm sorta in the middle of working some linear algebra stuff but I'll try to help in a few minutes if no one else comes to the rescue haha.

ganeshie8 · Answer

what i don't get is how the kinetic energy of gamma rays matters here
wavelength of gamma rays shouldn't matter on the kinetic energy right ?

kainui · Answer

Well gamma rays is sorta a range of energy, there's not just one single wavelength that is gamma radiation. It's exactly the same reason why there are multiple radio stations, even though they're all using radio waves.

ganeshie8 · Answer

what does it mean to say ` 100 keV gamma rays`

i know `100 keV electron` has a kinetic energy of `100keV`, but how to think of the same for gamma rays ?

kainui · Answer

There's really only one photon, but the photon can have different frequencies and polarizations. I think it becomes confusing because people think the speed of light is constant but frequency can change and that also seems to be like a speed. But the difference is like you're driving a car at a constant speed and the passenger is jumping up and down in their seat either faster or slower which is the frequency aspect of the photon.

ganeshie8 · Answer

for a photon of given frequency, the energy is fixed right ?

ganeshie8 · Answer

what does it mean to say ` 100 keV gamma rays` ?

kainui · Answer

They didn't have to say "gamma rays" they could have said 100 keV light and that would have been exactly the same. It's a little redundant but they're just trying to make you associate high energy with the waves in that region of the spectrum.

ganeshie8 · Answer

Ahh okay, gotcha!
so the photon has an energy 100keV
from this i can calculate the wavelength, nice :)

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=solve+100*10^3*1.602677*10^%28-19%29++%3D+6.62607004+*+10^%28-34%29+*299792458%2FL

kainui · Answer

Alright I guess we should look at the de Broglie wavelength. There are two sorta nice ways to memorize it as a clump of stuff.
$$p=\frac{h}{\lambda}$$
There's the clear analogous formula:
$$E=h \nu$$

And then there's also the fact that you probably know the Heisenberg uncertainty relation 
$$\Delta p \Delta x \ge \frac{\hbar}{2}$$
Knowing the units here, you can immediately feel assured that you have the debroglie relation correct since $\lambda$ is wavelength and has units of distance and so does $\Delta x$ by comparison.

In spectroscopy we tend to use wavenumbers which are to wavelength as what frequency is to period, and that makes the equations look much more symmetric looking. We define wavenumbers as: $$\tilde \nu = \frac{1}{\lambda} $$
Also, I'll include the angular versions since I think they are found a lot as well.

$$p=h \tilde \nu = \hbar k$$$$E=h \nu=\hbar \omega$$

ganeshie8 · Answer

$\lambda \approx 12.4pm$ looks good ?

kainui · Answer

That sounds pretty good, since gamma rays are high energy, they should have very short wavelength.

ganeshie8 · Answer

defining $1/\lambda$ by a different variable looks nice...

ganeshie8 · Answer

then the electron must be accelerated a kinetic energy such that its wavelength becomes $12.4pm$

kainui · Answer

Yeah, normally it's written that way it means it has units of inverse centimeters. Literally you can understand wavenumbers as "number of waves per centimeter" since wavelength is normally "centermeters per length of wave".

Sorry for the delay I got sidestepped thinking about the photoelectric effect but I realized we're using voltage to eject electrons not photons haha.

ganeshie8 · Answer

Nice, we're using voltage to accelerate already existing electrons i think...

kainui · Answer

Well alright so let's continue on, energy is just like we'd expect for an electron classically. It's on the plate and we gotta accelerate it off. So it has no kinetic energy from the start and it ends up with kinetic energy. There's gonna be a change in potential energy which is proportional to the change in voltage, since voltage is joules per charge.

$$U_i+T_i = U_f+T_f$$ I just explained :$$T_i=0$$ and $$U_f-U_i = q\Delta V$$ So we have:

$$q\Delta V = T_f = \frac{p^2}{2m}$$

ganeshie8 · Answer

can i use the classical equation
p^2/2m = KE 
?

ganeshie8 · Answer

so when the electron goes through that potential difference, it gets the required kinetic energy

ganeshie8 · Answer

il plugin the numbers, one sec..

kainui · Answer

I hope it works, I'm gonna be right back real fast gonna make some tea.

ganeshie8 · Answer

$p = 5.34*10^{-22}$ http://www.wolframalpha.com/input/?i=%286.62607004+*+10^%28-34%29%29%2F%281.24*10^%28-12%29%29

ganeshie8 · Answer

$$\Delta V = \dfrac{p^2}{2em} \approx 977kV$$ http://www.wolframalpha.com/input/?i=%285.34*10^%28-22%29%29^2%2F%282*1.60217662*10^%28-19%29*9.10938356+*+10^%28-31%29++%29

kainui · Answer

I'm getting roughly 3.89 pm for the wavelength: http://www.wolframalpha.com/input/?i=(plancks+constant)%2Fsqrt(2*(mass+of+electron)*100keV)

kainui · Answer

Whoops I've done it backwards haha. I see what I've done, I plugged in the energy of the gamma ray looking for the wavelength but I need to find the energy given that the wavelength is the same haha.

kainui · Answer

No wonder I felt so backwards in plugging it in, so to clarify the general flow I should have gone through (which I'm just gonna assume you've done correctly) is:

energy of photon -> wavelength of photon -> wavelength of electron -> energy of electron

ganeshie8 · Answer

hey should i treat 100keV as kinetic energy K = p^2/2m or the actual energy E = hf ?

ganeshie8 · Answer

correct answer is 9.76 kV hmm

ganeshie8 · Answer

im missing a 100 somewhere in my calculation..

kainui · Answer

Here, I think I confused you. You should treat it as E=hf since it's the energy of light.

A microscope's resolving ability relies on the wavelength of the particles you shoot at it, whether they be photons or electrons. So first off, I'll call light energy as $E_\ell$ to distinguish it from the electron energy $E_e$.

$$E_\ell = \frac{h c}{\lambda_\ell}$$
In order to maintain that both the microscopes have the same resolving ability, $\lambda_\ell = \lambda_e$ so now we can start to solve for $E_e$ in terms of $E_\ell$. From now on I'll just refer to the wavelength as $\lambda$ since they are identical, that's the point of this problem.

$$E_e= \frac{p^2}{2m} = \frac{h^2}{2m \lambda^2 }$$
We can get $\frac{h^2}{\lambda^2}$ from the light energy equation to make it a breeze to plug in.

$$\frac{E_\ell^2}{ c^2} = \frac{h^2}{\lambda^2}$$ 
$$E_e = \frac{E_\ell^2}{2mc^2}$$

The takeaway from this problem is that $c^2$ in the denominator is a huge number so correspondingly the energy required to make an electron microscope of the same resolving ability as a light microscope is MUCH less.

kainui · Answer

http://www.wolframalpha.com/input/?i=100keV%2F(2*(mass+of+electron)*(speed+of+light)%5E2)&t=crmtb01 Yeah I think I am missing some powers of 10s somewhere

kainui · Answer

Well, the weird thing is that the answer WA gives me has no units attached wtf

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=%28100*10^3*1.602677*10^%28-19%29%29%2F+%282*9.10938356+*+10^%28-31%29*+299792458^2%29

ganeshie8 · Answer

by that reasoning electron microscope is better than the corresponding light microscope of same resolvability

ganeshie8 · Answer

your derivation looks fine to me, il check why we are not getting the textbook answer...

kainui · Answer

In practice, that's not exactly true shooting photons and shooting electrons at things can have some pretty different results. But yeah electron microscopes are awesome lol.

ganeshie8 · Answer

can we really use gamma rays in microscope to see fine details of say some metal surface ?

inkyvoyd · Answer

Of course. How do you think we figured out the structure of DNA?

kainui · Answer

I actually really love the structure of DNA's from x-rays it has such a beautiful explanation. But x-rays are weaker than gamma rays, and I don't know about gamma rays since I only worked in an x-ray lab. I think gamma rays are so high energy it would be impractical though @ganeshie8

ganeshie8 · Answer

interesting.. so ppl are using gamma rays in microscopes is it, nice..

inkyvoyd · Answer

what kainui said is true; x-rays are much softer than gamma rays, so they are not one to one, but they turn out to be enough...

ganeshie8 · Answer

are you talking about x ray diffraction ?

inkyvoyd · Answer

x-ray crystallography actually, which is similar

inkyvoyd · Answer

applied x-ray diffraction

kainui · Answer

Ok guys I think you really need to learn this because this is probaly the most amazing thing I've ever learned when I was working in an x-ray lab. https://upload.wikimedia.org/wikipedia/en/b/b2/Photo_51_x-ray_diffraction_image.jpg and now look at the spring on the far left. http://www.yamahapart.com/Assets/YamahaPart/Images/workshop/suspension_principles/6_1_1_3_1_PS_gc_single_col.jpg Now if you see in the spring there is \\\ and ////// These are the diffraction gratings of DNA's helix that forms the X pattern in the first picture.

ganeshie8 · Answer

my textbook says we need to use some crystal structures to see the diffraction phenomenon of xrays as we can't physically make apertures small enough for xrays to diffract..

ganeshie8 · Answer

if i understand correctly, in xray lab, you ppl are using DNA structure itself as a diffraction grating for xrays to diffract ?

ganeshie8 · Answer

by studying the distance between interference fringes etc you're guessing the actual structure of the DNA.... fascinating!

idk how much nonsense i have talked in past two replies based on my bookish knowledge lol

kainui · Answer

Yeah, the zigzag you see from the profile of a spring is a pair of diffraction gratings and that's what we, human beings, did to figure out that we're made up of these little helices.

ganeshie8 · Answer

its so cool !
can u recommend any document that explains xray diffraction in a simple manner ?

ganeshie8 · Answer

i mean the applications of xray diffraction, like, to study crystal structures etc... my textbook really did a good job explaining the basics of xray diffraction, but it didn't bother to go through any applications..

kainui · Answer

Hmmm well you can study solid state chemistry from a physical chemistry textbook. I guess I should tell you, spectroscopy is the study of the interaction of light and matter and has like a billion applications in ways you had no idea even existed.

Maybe look into Bragg's law? I don't have anything in particular off hand, I have some textbooks but I wouldn't suggest buying them.

ganeshie8 · Answer

Bragg's law is about polarization by reflection right ? My textbook has covered it already

ganeshie8 · Answer

it does a good job of explaining the physics, but it rarely mentions about the practical applications

kainui · Answer

x-ray diffraction is one application of it, you shoot xrays at a crystal, move the machine through an angle recording if it absorbed or not and you get peaks which correspond to looking at the different views which line up with the crystal lattice. Bravais lattices, Miller indices, and Fourier analysis is all part of this, maybe look into that as well?

ganeshie8 · Answer

Okay thanks! math behind these doesn't look pleasant haha
 I'll look up them when free :) i guess for now il continue w/ my QM :)

kainui · Answer

Yeah keep with it, work through the pain lol.

ganeshie8 · Answer

$$E_e = \frac{E_\ell^2}{2mc^2}$$ http://www.wolframalpha.com/input/?i=%28100*10^3*1.602677*10^%28-19%29%29^2%2F+%282*9.10938356+*+10^%28-31%29*+299792458^2*1.60217662+*+10^%28-19%29%29

thomas5267 · Answer

Is x ray diffraction that hard?  Furthermore, what is happening here?  The calculation should be relatively simple.  1 eV is the KE of a electron when accelerated through 1 V.  A 100 eV gamma ray has the energy of the kinetic energy of electron accelerated through 100 V.  Use the de Broglie wavelength formula (also known as TheBroccoli wavelength by my friend) to find the momentum needed for the electron to have the same wavelength as the photon, then calculate the kinetic energy using $\dfrac{p^2}{2m}$.  Finally change the unit back to eV if necessary.

kainui · Answer

lol