The nucleus of an atom consists of protons and neutrons (no electrons). a nucleus of a carbon‑12 isotope contains six protons and six neutrons, while a nitrogen‑14 nucleus comprises seven protons and seven neutrons. a graduate student performs a nuclear physics experiment in which she bombards nitrogen‑14 nuclei with very high speed carbon‑12 nuclei emerging from a particle accelerator. as a result of each such collision, the two nuclei disintegrate completely, and a mix of different particles are emitted, including electrons, protons, antiprotons (with electric charge − each), positrons (with charge + each), and various neutral particles (including neutrons and neutrinos). for a particular collision, she detects the emitted products and find 17 protons, 4 antiprotons, 7 positrons, and 25 neutral particles. how many electrons are also emitted?

7 electrons

Explanation:

We can solve the problem by using the law of conservation of electric charge: in fact, the total electric charge before and after the collision must be conserved.

Before the collision, we have:

- A nucleus of carbon-12, consisting of 6 protons (charge +1 each) + 6 neutrons (charge 0 each), so total charge of +6

- A nucleus of nitrogen-14, consisting of 7 protons (charge +1 each) + 7 neutrons (charge 0 each), so total charge of +7

So the total charge before the collision is +6+7=+13 (1)

After the collision, we have:

- 17 protons (charge +1 each): total charge of +17

- 4 antiprotons (charge -1 each): total charge of -4

- 7 positrons (charge +1 each): total charge of +7

- 25 neutral particles (charge 0 each): total charge of 0

- N electrons (charge -1 each): total charge of -N

So the total charge after the collision is +17-4+7+0-N=+20-N (2)

Since the charge must be conserved, we have (1) = (2):

+13 = +20 - N

Solving for N,

N = 20 - 13 = 7

So, there are 7 electrons.

snell's law

refractve index = sin incidence / sin refraction