That's roughly equivalent to looking at an object 250nm wide at arm's length. A red blood cell is approximately 8000nm wide.
Crazy resolving power.
[1] http://blogs.discovermagazine.com/badastronomy/2010/01/12/sp...
ALMA consists of 66 antennas, most of which are 12 meters in diameter. That's about 7000 square meters of receiving area.
Betelgeuse is 642 light years away, which is 6x10^18 meters. The area of a sphere with that diameter is about 10^38 square meters. So 10^-34 of the power emitted from Betelgeuse ends up falling on the ALMA array.
According to Wikipedia, the luminosity of Betelgeuse is 90-150 thousand solar luminosity units, which is about 4x10^26 watts. Let's call it 10^31 watts. So the total power received from Betelgeuse by ALMA is about a milliwatt.
But that's the total power, and the ALMA array only receives at 0.32 to 3.6 mm. To figure out what proportion of Betelgeuse's power falls in this range we need to integrate over Betelgeuse's spectrum, both the total spectrum and then this range in order to find the ratio. That part of the calculation is not so easy. But let's see what we can do. Let's assume that Betelgeuse has a blackbody spectrum. Its temperature is 3500K. We can use this handy dandy blackbody spectrum calculator:
http://www.spectralcalc.com/blackbody_calculator/blackbody.p...
When you crunch the numbers it turns out that about 10^-7 of the total power falls in the range 0.32 to 3.6mm. So the total power received by ALMA is about 10^-10 watts.
0.32-3.6mm is in the far infrared. A photon at this wavelength has an energy of about one meV, or about 10^-22 Joules. So 10^-10 watts is about 10^12 photons per second.
I don't know how long the exposure times are, but my guess is that they are measured in hours.