2
$\begingroup$

I notice that Wilson invented his could chamber in 1911 and Hess discovered cosmic rays in 1912. So, did Wilson use his chamber to study cosmic rays or did Hess do so?

Today, it is easy to detect cosmic rays with a home-made cloud chamber. Hence, Wilson should have detected them with his chamber, right?

$\endgroup$
  • $\begingroup$ Minor quibble: homemade cloud chambers normally detect secondary particles produced by cosmic rays, as I understand. $\endgroup$ – Emilio Pisanty Mar 3 '17 at 13:14
3
$\begingroup$

Wilson started to develop the Cloud Chamber in the late 19th to reproduce the optical glory. He soon recognized it could be used to detect particles. In 1911 he perfected his device to take pictures and started tests with X-rays and beta particles.

During his experiments there were some unwanted ion productions in the chamber even with no source of radiation. Wilson suggested that that may be due to extreme radiation coming through the atmosphere, what would be the first idea of cosmic rays. He proceeded with tests in "isolated environments" such as tunnels and the effected did not change so he abandoned this idea.

In Wilson's Nobel Lecture he does not mention about cosmic rays so it might be possible he worked only with X-rays, alpha and beta particles at least in the early days of the Cloud Chamber.

$\endgroup$
2
$\begingroup$

The Austrian Nobel price winner (1936) Viktor F. Hess discovered the cosmic radiation during several ballon flights in 1912 with specially designed ionization chambers. He discovered that (unexpectedly) the penetrating radiation increased with altitude and concluded that there must be a extraterrestrial source for this radiation.(Hess paper 2012) Wilson's cloud chamber made the tracks of ionized particles visible. He did not use his chamber to study cosmic radiation which was unknown to him in 2011. The cloud chamber detects ionized particles which are mainly produced by terrestrial radioactive elements at sea level.

$\endgroup$
  • $\begingroup$ "mainly produced by terrestrial radioactive elements at sea level." is wrong, there is one cosmic ray muon passing through a cm^2 area each minute at sea level hyperphysics.phy-astr.gsu.edu/hbase/Particles/muonatm.html. Considering that cloud chambers will have crossections more than 60cm^2 one should expect quite a few in a cloud chamber per second. $\endgroup$ – anna v Mar 4 '17 at 7:05
  • $\begingroup$ Was testing a prototype spark chamber with cosmic muons back in 1965. Radioactive elements are easy to be separated because of thicker tracks in a cloud chamber, and because the source can be moved. One uses anticoincidence to avoid the cosmic muons in sophisticated experiments. $\endgroup$ – anna v Mar 4 '17 at 7:28
  • $\begingroup$ @anna v - According to this article of the International Atomic Energy Agency (IAEA) the annual radiation dose from natural terrestrial radioactive sources (2.0 mSv, 84%) is large as compared to the annual radiation dose due cosmic radiation (0.4 mSv, 16%): iaea.org/sites/default/files/publications/magazines/bulletin/… $\endgroup$ – freecharly Mar 4 '17 at 21:27
  • $\begingroup$ Most of it comes from materials like walls (granite) and floor cracks (radon) . I am just commenting of the incident cosmic muons which are enough to appear in a cloud chamber, and are very noticeable and different ( muons long straight tracks, alphas betas short fat ones) . What people absorb is a different measure. $\endgroup$ – anna v Mar 5 '17 at 4:20
  • $\begingroup$ @anna v - You are , of course, right that cosmic rays are also detected in the radiation background at sea level. I think that Viktor Hess measured mostly X-rays in his ionization chambers which produce energetic electrons and ionized molecules/atoms which generate the measured electron/ion pairs along their tracks in the chamber. Such x-rays are also produced in the decay series of different terrestrial radioactive isotopes. The ion production (absorbed energy) in a gas by these rays is probably similar to the absorbed energy in other materials. $\endgroup$ – freecharly Mar 5 '17 at 5:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.