1.What is the particle xxx in the following nuclear reaction: 49Be+24He→612C+x^{9}_{4}\text{Be} + ^{4}_{2}\text{He} \to ^{12}_{6}\text{C} + x49Be+24He→612C+xa.electronb.protonc.photond.neutronLogin to continueOnly logged in users canattempt or see the solution.
