Exercise practice: Structure of Atom

6.1 How the exercises are grouped

The chapter prints exercises 2.1 to 2.67. They test five skills, and the worked questions below cover the method for each group. The group table names every exercise the method applies to.

6.2 Group A: counting particles and their charge

Exercise 2.1 asks how many electrons weigh one gram. Divide 1 g by the electron mass in grams, 9.109 x 10^-28 g, which gives 1.098 x 10^27 electrons. One mole of electrons has mass 6.022 x 10^23 x 9.109 x 10^-31 kg, which is 5.485 x 10^-4 kg, and its charge is 6.022 x 10^23 x 1.602 x 10^-19 C, which is 9.648 x 10^4 C.

Exercise 2.3 asks for protons and neutrons in five nuclei. The proton count is the atomic number and the neutron count is the mass number minus the atomic number. Carbon-13 has 6 protons and 7 neutrons, oxygen-16 has 8 and 8, magnesium-24 has 12 and 12, iron-56 has 26 and 30, and strontium-88 has 38 and 50. Exercise 2.4 reverses the task: an element with Z = 17 and A = 35 is chlorine-35, and one with Z = 92 and A = 233 is uranium-233.

Exercise 2.38 gives a static charge of 2.5 x 10^-16 C and asks for the electron count. The charge is a whole number multiple of the elementary charge, so divide 2.5 x 10^-16 by 1.602 x 10^-19, giving 1560 electrons.

6.3 Group B: photons and spectra

Exercise 2.22 belongs here only for its isoelectronic counting idea, but the photon exercises follow one pattern. Convert the wavelength to metres, then use the photon relations. If the wavelength is 600 nm, the frequency is c divided by the wavelength, which is 5.0 x 10^14 Hz, and the photon energy is h nu, which is 3.3 x 10^-19 J or 2.1 eV.

Exercise 2.7 gives a period of 2.0 x 10^-10 s. Frequency is the reciprocal of the period, so nu is 5.0 x 10^9 Hz. The wavelength is c divided by nu, which is 6.0 x 10^-2 m, and the wavenumber is 1 over the wavelength, 16.7 m^-1.

Exercise 2.53 uses a stopping voltage of 0.35 V with radiation of wavelength 256.7 nm. The photon energy is 1240 divided by 256.7, which is 4.83 eV. The work function is the photon energy minus the stopping energy, 4.83 minus 0.35, which is 4.48 eV. This is the energy needed to free an electron from silver.

Exercises 2.55 and 2.56 use the Rydberg relation in reverse. Find the wavenumber, convert to wavelength, and then compare with the series limits to name the series. In 2.56 a transition from a radius of 1.3225 nm to 211.6 pm belongs to the Balmer series because the lower orbit with radius 211.6 pm is n = 2.

6.4 Group C: the Bohr model

Exercise 2.16 uses the two central formulas. The fifth orbit of hydrogen has energy -13.6/25, which is -0.544 eV, and radius 52.9 x 25, which is 1322.5 pm or 1.3225 nm. Exercise 2.19 removes the electron completely from n = 2, which needs 13.6/4, that is 3.4 eV. The longest wavelength that can do this is hc divided by that energy, 1240/3.4, which is 365 nm.

Exercise 2.17 finds the wavenumber of the longest wavelength Balmer line. The longest line is the smallest energy step, the 3 to 2 transition. The wavenumber is R(1/4 minus 1/9), which is 1.097 x 10^7 x 5/36, giving 1.52 x 10^6 m^-1.

Exercise 2.34 asks for the energy to remove the electron from He+. The hydrogen ionisation energy is 2.18 x 10^-18 J, and the nuclear charge of helium doubles the factor Z squared, so the energy is four times larger, 8.72 x 10^-18 J per atom.

Exercise 2.32 shows that the circumference of the nth Bohr orbit is n times the de Broglie wavelength. Use 2 pi r equal to n lambda, substitute the Bohr radius and the de Broglie wavelength, and the relation reduces to the angular momentum condition.

6.5 Group D: matter waves and uncertainty

Exercise 2.20 gives an electron speed of 2.05 x 10^7 m s^-1. The momentum is 9.109 x 10^-31 times 2.05 x 10^7, which is 1.87 x 10^-23 kg m s^-1, and the de Broglie wavelength is 6.626 x 10^-34 divided by that momentum, which is 3.55 x 10^-11 m.

Exercise 2.21 starts from kinetic energy. The de Broglie wavelength can be written as h divided by the square root of 2mK. Substituting K equal to 3.0 x 10^-25 J gives 8.97 x 10^-7 m.

Exercise 2.59 uses the speed of the electron in the first Bohr orbit, 2.19 x 10^6 m s^-1. The wavelength is 6.626 x 10^-34 divided by (9.109 x 10^-31 x 2.19 x 10^6), which is 3.32 x 10^-10 m, about one third of a nanometre.

Exercise 2.61 fixes the position uncertainty at 0.002 nm. The momentum uncertainty is h divided by 4 pi times the position uncertainty, and the large value it gives shows why a definite path for the electron cannot be defined.

6.6 Group E: quantum numbers and configurations

Exercise 2.24 asks for the lowest n that allows g orbitals. The letter g corresponds to l = 4, and l cannot exceed n minus 1, so the smallest possible n is 5. Exercise 2.25 says an electron is in a 3d orbital, so n = 3 and l = 2, and m_l can be -2, -1, 0, +1 or +2.

Exercise 2.28 lists n = 3 and asks for l and m_l. For n = 3, l can be 0, 1 or 2. For l = 2 the m_l values are -2 to +2. Among the listed orbitals, 1p and 3f are impossible because l cannot reach 1 when n is 1, and cannot reach 3 when n is 3. Exercise 2.30 rejects sets that break the limits: n cannot be 0, l cannot equal n, and m_l cannot lie outside the range from minus l to plus l.

Exercise 2.22 checks isoelectronic species. Sodium ion and magnesium ion each have 10 electrons. The potassium ion, calcium ion, sulphide ion and argon atom each have 18 electrons, so those four are isoelectronic with one another.

Exercise 2.67 asks about n = 4. There are four subshells, 4s, 4p, 4d and 4f, and a total of 16 orbitals. Half of the 32 electrons in the shell have m_s = -1/2, so the count is 16.