The Nuclear Model of an atom Worked ExamplesIGCSE Physics 0625

(a) The nucleus contains two kinds of particle: J and K. M orbits the nucleus, so M is an electron (relative charge $-1$). There are three electrons, and there are also three J particles, so J must be the proton (relative charge $+1$) — a neutral atom has equal numbers of protons and electrons. K is therefore the neutron (relative charge $0$).

(b) Count the particles in the nucleus. Proton number $Z = 3$. Nucleon number $A$ = protons + neutrons.

(c) The atom has $3$ protons (total charge $+3$) and $3$ electrons (total charge $-3$). Neutrons are uncharged, so the total charge is zero.

From the nuclide notation: $Z = 9$ protons, and neutrons $= A - Z$.

The neutral atom has $9$ electrons. Forming an ion changes only the electrons — the nucleus is unchanged. Gaining one electron gives $10$ electrons.

Charge $= 9 \times (+1) + 10 \times (-1) = -1$. This is row B.

A describes losing an electron (a positive ion). C changes the nucleus, which does not happen when an ion forms. D uses the nucleon number $19$ as the number of neutrons.

(a) Isotopes are atoms of the same element — so they have the same number of protons (same proton number) — but different numbers of neutrons (different nucleon numbers).

(b) Both have $Z = 17$, so $17$ protons and (as neutral atoms) $17$ electrons. Neutrons $= A - Z$:

(c) The relative charge of a nucleus equals its proton number (each proton $+1$, neutrons $0$): both nuclei have relative charge $+17$.

The relative mass of a nucleus equals its nucleon number (protons and neutrons each have a relative mass of about $1$): $35$ and $37$.

For an alpha particle to be turned right round and sent back, it must hit something much more massive than itself, concentrated in a tiny space. So the rebounding particles show that the nucleus contains most of the mass of the atom: C.

A is evidence that the atom is mostly empty space (with a very small nucleus). B shows that the alpha particles are repelled, so the nucleus is positively charged.

D is a feature of how the experiment is set up, not an observation of the results.

(a) Most alpha particles pass through with little or no deflection, so the atom is mostly empty space — the nucleus is very small compared with the atom.

(b) Alpha particles are positively charged. Q passes close to the nucleus and is repelled (pushed away from it), so the nucleus must be positively charged — like charges repel.

(c) R is turned back, so the nucleus must be much more massive than an alpha particle: the nucleus contains most of the mass of the atom.

Because hardly any alpha particles come close enough to be turned back, the nucleus must be a very small target — its mass is concentrated in a tiny volume.

(a) Read the bars: $16$ protons, $18$ neutrons, $18$ electrons.

(b) Proton number $16$ is sulfur.

(c) Add the charges: protons $+1$ each, electrons $-1$ each.

There are $2$ more electrons than protons, so Y is a negative ion — a sulfur atom that has gained two electrons.

(d) The nucleus contains only the protons and neutrons, so its relative charge is $+16$ — the electrons are outside the nucleus.

(a) Neutrons $= A - Z$ for each.

(b) Both nuclei have $53$ protons. Iodine-131 has four more neutrons than the stable isotope.

It has too many neutrons compared with its number of protons (excess neutrons), so its nucleus is unstable. An unstable nucleus is radioactive — it changes to become more stable by emitting radiation.

(a) Nucleon numbers must balance on both sides of the equation.

Proton numbers (charge) must also balance.

(b) A particle with nucleon number $1$ and proton number $0$ (no charge) is a neutron, ${}^{1}_{0}\text{n}$. (A proton would be ${}^{1}_{1}\text{p}$ and a beta particle ${}^{0}_{-1}\text{e}$.)

(a) Two light nuclei join to form a heavier nucleus: this is nuclear fusion.

(b) Balance the nucleon numbers and the proton numbers.

So X is a neutron, ${}^{1}_{0}\text{n}$.

(c) The total mass of the products is slightly less than the total mass of the two deuterium nuclei. This missing mass has been converted into energy, which is released (as kinetic energy of the products).

(a) Isotopes have the same proton number, so they lie in the same vertical column: A and B ($Z = 6$); C and F ($Z = 7$); D and E ($Z = 8$).

(b) Nucleon number $A = Z + N$. Calculate it for each nucleus.

So B and D have the same nucleon number as C ($14$). (They are different elements with the same nucleon number — not isotopes.)

(c) Relative charge of a nucleus = proton number. The largest is $Z = 8$: nuclei D and E, relative charge $+8$.

(d) Relative mass of a nucleus = nucleon number. The largest is $16$: nuclei E and F.

(e) F has $Z = 7$ and $N = 9$, so $A = 16$.

(a) A heavy nucleus splits into two smaller nuclei: nuclear fission.

(b) Balance nucleon numbers. Remember that two neutrons are produced, so they contribute $2 \times 1 = 2$ on the right.

Balance proton numbers (neutrons have $Z = 0$).

$Z = 52$ is tellurium, so X is ${}^{137}_{52}\text{Te}$.

(c) The total mass of the products (the two fragments and the neutrons) is slightly less than the total mass of the uranium nucleus and the neutron before. The lost mass is converted into energy, released mainly as kinetic energy of the fragments and neutrons.

(a) Alpha particles are absorbed by a few centimetres of air (they collide with air molecules). Removing the air lets them reach the foil and the detector.

(b) A thick foil would absorb the alpha particles, and each particle would be deflected by many nuclei in turn. A very thin foil means most particles pass through and each is scattered by at most one nucleus, so the results can be interpreted.

(c) The count rate decreases very rapidly as $\theta$ increases: most alpha particles are deflected through small angles. Compare the extremes:

(d) Some alpha particles are scattered through $165^{\circ}$ — almost straight back — so the nucleus is positively charged and contains most of the mass of the atom. Because so few are scattered this far, the nucleus must be very small (most of the atom is empty space).

(e) At $165^{\circ}$ only about $0.057 \times 60 \approx 3$ alpha particles arrive per hour. Emission and scattering are random, so a long counting time is needed to record enough counts for a reliable average.