4 ms·

Why does the atom look so big in the bicture? Atomic radius of Strontium is 219 pm, so that small spec there in the picture should be about 438 pm across. I'm

Why does the atom look so big in the bicture? Atomic radius of Strontium is 219 pm, so that small spec there in the picture should be about 438 pm across. I'm assuming the two ball-pen nib shaped structures on both sides of the spec in the picture are "two metal electrodes placed about 2mm (0.078in) apart". So, the space between the left tip of the electrode and left edge of the spec is about 1 mm. Based on some "visual calculation" (zooming in the picture and doing some approximation), the spec seems to be closer to about 0.03 mm across, which is orders of magnitude larger than 438 pm that it should be. What gives?

tjohns 9y ago
You're not looking directly at the atom itself (which is impossible... atoms are smaller than light's wavelength). You're looking at the photons emitted by an excited atom, as collected over an extremely long exposure by the camera's sensor. Which will resolve to ~1px in size... the smallest unit the camera can image.
- SeanLuke 9y ago
  It doesn't appear that the atom is 1 pixel, but is in fact quite a bit larger than this: there are lots of elements in the image which are significantly smaller than it. So what gives?
  - tjohns 9y ago
    Possibly diffraction, possibly JPEG artifacts, possibly camera movement. Tough to say without more details. Looking at it, my suspicion is diffraction. But I don't know enough optics to be sure.
  - gji 9y ago
    I used to work on trapped ion experiments, and the limitation of the atom size was always the diffraction limit, which is limited by the NA of the lens (f-stop in camera terms) and the wavelength of light. In this case, the optical system (I'm guessing a camera lens) is designed for multiple wavelengths, so it might not reach the diffraction limit at the emission wavelength, which is like 400nm. In that case, the limit would be the aberration of the camera lens, which can be wavelength-dependent. Most camera lenses aren't designed for 400nm light, which is marginally visible.
- StavrosK 9y ago
  It doesn't really mean anything to day "you aren't looking at it, you're looking at the reflected photons". You're always looking at reflected photons of everything.
  - kgwgk 9y ago
    Actually “reflected” doesn´t mean much for photons: they are absorbed and re-emitted.
    - StavrosK 9y ago
      Isn't that what "reflected" means?
      - kgwgk 9y ago
        Not exactly. https://en.m.wikipedia.org/wiki/Reflection_(physics) https://en.m.wikipedia.org/wiki/Reflection_(physics) For most of the history of the word “reflect”, its meaning was definitely not “absorb and re-emit photons of the same frequency”.
SAI_Peregrinus 9y ago
The atom can't be directly imaged by bouncing light off it, it's far too small to do that (with visible light). Instead they're shining a laser at it, which excites some of its valence electrons to higher orbitals. When those electrons drop back to their ground states they emit (visible) light, some of which reaches the camera. The atom is effectively acting as an isotropic radiator (radiating equally in all directions). The camera lens is much larger than the atom. There are thus multiple paths from the atom to the lens and the sensor behind. Less light will reach the camera from greater angles, so the light that goes straight towards the lens or nearly straight will impact the sensor most, and that area will appear brightest. (This is my somewhat mangled attempt to explain abberation in optics...) Even if the light were perfectly collimated in a beam the size of the atom it would still be unable to make a dot in the final image smaller than a single pixel of the sensor. If film were used instead of a digital sensor it would expose at least one pigment grain, again much larger than the atom itself. The apparent size is an artifact of the imaging process.

tjohns 9y ago

You're not looking directly at the atom itself (which is impossible... atoms are smaller than light's wavelength). You're looking at the photons emitted by an excited atom, as collected over an extremely long exposure by the camera's sensor. Which will resolve to ~1px in size... the smallest unit the camera can image.

SeanLuke 9y ago
It doesn't appear that the atom is 1 pixel, but is in fact quite a bit larger than this: there are lots of elements in the image which are significantly smaller than it. So what gives?
- tjohns 9y ago
  Possibly diffraction, possibly JPEG artifacts, possibly camera movement. Tough to say without more details. Looking at it, my suspicion is diffraction. But I don't know enough optics to be sure.
- gji 9y ago
  I used to work on trapped ion experiments, and the limitation of the atom size was always the diffraction limit, which is limited by the NA of the lens (f-stop in camera terms) and the wavelength of light. In this case, the optical system (I'm guessing a camera lens) is designed for multiple wavelengths, so it might not reach the diffraction limit at the emission wavelength, which is like 400nm. In that case, the limit would be the aberration of the camera lens, which can be wavelength-dependent. Most camera lenses aren't designed for 400nm light, which is marginally visible.
StavrosK 9y ago
It doesn't really mean anything to day "you aren't looking at it, you're looking at the reflected photons". You're always looking at reflected photons of everything.
- kgwgk 9y ago
  Actually “reflected” doesn´t mean much for photons: they are absorbed and re-emitted.
  - StavrosK 9y ago
    Isn't that what "reflected" means?
    - kgwgk 9y ago
      Not exactly. https://en.m.wikipedia.org/wiki/Reflection_(physics) https://en.m.wikipedia.org/wiki/Reflection_(physics) For most of the history of the word “reflect”, its meaning was definitely not “absorb and re-emit photons of the same frequency”.

SAI_Peregrinus 9y ago

The atom can't be directly imaged by bouncing light off it, it's far too small to do that (with visible light). Instead they're shining a laser at it, which excites some of its valence electrons to higher orbitals. When those electrons drop back to their ground states they emit (visible) light, some of which reaches the camera. The atom is effectively acting as an isotropic radiator (radiating equally in all directions). The camera lens is much larger than the atom. There are thus multiple paths from the atom to the lens and the sensor behind. Less light will reach the camera from greater angles, so the light that goes straight towards the lens or nearly straight will impact the sensor most, and that area will appear brightest. (This is my somewhat mangled attempt to explain abberation in optics...) Even if the light were perfectly collimated in a beam the size of the atom it would still be unable to make a dot in the final image smaller than a single pixel of the sensor. If film were used instead of a digital sensor it would expose at least one pigment grain, again much larger than the atom itself. The apparent size is an artifact of the imaging process.