James Chadwik found the neutron by shoot α rays to a Beralium foil. α rays are equal to the He²⁺ and he saw that particles without a charge is out from that. He understood the neutrons are those particles without the charge.

But my question is this.

This is the equation that the chadwik found.

Be + He²⁺= C +neutron

In this equation the left side has a 2+charge, but in right there is no any charge. So, What happen to the 2+ charge when this reaction occurs?

  • 3
    $\begingroup$ For nuclear reactions, usually occurring at very high energies (compared with typical electron binding energies) one just ignores the electrons - they are going to do what they will do, and have zero impact on what the nucleons are doing. So, that would usually be written $^{9}Be (\alpha, n) ^{12}C$. The alpha was just a way to get an energetic-enough alpha to enter the nucleus and react. $\endgroup$
    – Jon Custer
    Aug 15 '17 at 15:36
  • $\begingroup$ Electrons have no importance whatsoever in the reaction. $\endgroup$
    – Mithoron
    Aug 15 '17 at 17:11
  • $\begingroup$ When $\ce{He^{2+}}$ ions hit a beryllium target, then the target will become charged. However beryllium is an excellent conductor. So if you electrically ground the target then the charge will disparate. // You have to realize that energetic $\ce{He^{2+}}$ ions will leave a trail of ions as the $\ce{He^{2+}}$ travels through the beryllium. This isn't like a chemical reaction. Also remember that you got the $\ce{He^{2+}}$ ion by stripping off two electrons. Those electrons are floating around somewhere in the universe... $\endgroup$
    – MaxW
    Aug 16 '17 at 3:27

Corrected: "James Chadwik identified the neutron by exposing a Beryllium foil to α-rays (He+2) and showing that the previously discovered emitted collision product were massive neutrally charged particles." The experimental set-up he used (see:http://rspa.royalsocietypublishing.org/content/royprsa/136/830/692.full.pdf) did not determine where any electrons came from, so while electrical charge can be assumed to balance, he couldn't claim any direct knowledge of the source or fate of any electrons in his experiment. For us, today, balancing a nuclear reaction's charge distribution on both sides of the arrow is simply "bookkeeping", it implies facts that were not as well supported in 1932 as they are today. He avoided the temptation to over-interpret his experimental data.


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.