For on-line help with this topic, see the chemcal module "Atomic and Nuclear Structure" which deals with the topics: Fundamental concepts; nuclear equations; electromagnetic radiation; wave-particle duality; Bohr model.

More advanced topics will be found in the following chemcal modules:

"Atoms , Electrons and Orbitals" which deals with the topics: Atomic orbitals - shapes, quantum numbers; subshell structures of atoms and ions.

"Atomic Properties" which deals with the topics: The concept of core charge and its relationship to fundamental atomic properties.

"Electronic Strudture of Atoms and Ions" which deals witht the topics: Trends in atomic properties in relation to the Periodic Table.

"Quantum Numbers and Atomic Orbitals" which is an advanced module on atomic orbitals.


Atomic structure (Advanced questions)

 

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Shortcut to Questions

Q: 1 2 3 4 5 6 7 8 9 10

1

For one atom of the isotope 34X (atomic number 16) give

(1) the number of protons

(2) the number of electrons

(3) the electronic configuration of the atom in the ground state

(4) an equation for one typical reaction of the element

2

Two isotopes of calcium (atomic number 20) are 40Ca and 44Ca.

For both 40Ca and 44Ca2+ supply the following information:

(1) atomic number

(2) mass number

(3) number of electrons

(4) number of neutrons

(5) number of protons

(6) electronic orbital configuration

3

(1) Write the detailed electronic configurations (in terms of sub shells) for the atoms of the following elements:

(a) Mn

(b) Zn

(2) In view of these configurations, account for the fact that manganese has a greater number of oxidation states than zinc.

4

(1) Calculate the wavelength of a bullet (mass 7.0 g) moving with a velocity of 1.1 x 103 m s-1.

(2) What is the wavelength of an electron moving at the same velocity? Why is the wave character of matter more apparent with the electron than with the bullet?

5

When a photon strikes a metal surface a certain minimum energy is required to eject an electron from the metal. This minimum, or threshold, energy is known as the work function of the metal. Any energy in the original photon above this minimum is translated into kinetic energy for the ejected electron. The threshold wavelength for photoelectric emission from lithium, above which no electrons are emitted, is 520 nm. Calculate the velocity of electrons emitted as the result of absorption of light at 360 nm.

6

For radiation which consists of photons of energy 5.0 x 10-16 J, find:

(1) the wavelength in nm

(2) the wavenumber in cm-1

(3) the frequency in Hz

7

The spectrum of mercury vapour shows strong emission at the wave lengths of 165 nm, 254 nm, 365 nm, 546nm, and 735 nm. Calculate the energy, in kJ mol-1, associated with each of these transitions, and state in which region of the electromagnetic spectrum the lines lie.

8

The Bohr model of the hydrogen atom can be used to predict the energy of various electronic transitions within the hydrogen atom. In this model, electrons are assumed to exist within certain "orbits" (designated by the symbol n) each of which has an associated discrete energy. The energy of an electron in a particular orbit n = E(n)

= (-2.18 x 10-18 J) x (1 / n2), for the hydrogen atom only.

(1.) Calculate the wavenumber (cm-1), and wavelength (nm) of the first three lines in the Lyman Series of the atomic spectrum of hydrogen.

(2.) What is the limiting wavelength of the series?

(3.) What is the energy of the final state (n = 1) for all transitions in the series?

(4.) What is the ionisation potential of the hydrogen atom, given eV = hn , where e = electronic charge and V = potential in volts.

9

Calculate the de Broglie wavelength (nm) of:

(1) an electron travelling with a velocity of 1.2 x 109 cm s-1

(2) a golf ball of mass 170 g, travelling with a speed of 3.58 x 103 cm s-1

10

The intense yellow light from a sodium street lamp arises from an electron jump from the 3p to the 3s level.

The wavelength of the light is 590 nm (5.9 x 10-7 m). What is the energy for this transition in kJ mol-1?

 

Atomic structure (Advanced answers)

 

1

(1) The number of protons equals atomic number, which is 16.

(2) We assume here that the atom in question is neutral, since no charges have been explicitly mentioned. In a neutral atom the number of electrons equals the number of protons, which is 16.

(3) The electronic configuration is found by assigning all available electrons to orbitals, starting with 1s. The configuration for 16 electrons will be 1s2 2s2 2p6 3s2 3p4.

(4) Noting that the outermost orbital (the 3p orbital) is two electrons short of being full, we would expect a typical oxidation state of "X" to be -II. Using 3d orbitals, +IV and +VI oxidation states are also possible. We can identify this as sulfur, which undergoes a wide variety of reactions. Sulfur can react with metals to produce the S2- ion (-II oxidation state), eg reaction with strontium:

Sr(s) + S(s) → SrS(s)

Sulfur can also react with non-metals to produce compounds in +IV and +VI oxidation states, eg burning in oxygen:

S(s) + O2(g) → SO2(g)

2

(1) The atomic number is unique for each element, and is the same for all isotopes of an element since it does not depend on the number of neutrons present. Therefore, the atomic number for both isotopes is 20. Note that atomic number depends only on the number of protons present, so the charge on the 44Ca2+ ion (which arises from the loss of two electrons) does not alter this.

(2) The mass number is the number of (protons + neutrons) present (ie the sum of the main contributors of mass within the atom). For 40Ca this is 40. Note that the charge present on the 44Ca2+ ion does not affect the composition of the nucleus, and the mass number is 44.

(3) The number of electrons equals the number of protons in a neutral atom, so the number of electrons in 40Ca is 20. The 44Ca2+ isotope, bearing a +2 charge, is missing 2 electrons, and therefore contains only 18 electrons.

(4) The number of neutrons is found from the difference between the atomic mass (no. protons + neutrons) and the atomic number (no. protons). So for 40Ca this equals 40 - 20 = 20 neutrons. For 44Ca2+ this equals 44 - 20 = 24. This shows that isotopes arise from differing numbers of neutrons.

(5) This is equal to the atomic number, and is 20 for both isotopes.

(6) 40Ca : 1s2 2s2 2p6 3s2 3p6 4s2

44Ca2+ : 1s2 2s2 2p6 3s2 3p6

Note the loss of the two outer electrons in the Ca2+ ion.

3

(1) (a) Mn : 1s2 2s2 2p6 3s2 3p6 3d5 4s2

(b) Zn : 1s2 2s2 2p6 3s2 3p6 3d10 4s2

(2) Mn is a transition element with an incompletely filled 3d shell. The completely filled 3d shell in zinc leads to stability and removal of the 4s electrons with only a relatively low ionization energy to form the 2+ ion is the only common oxidation state of zinc (+II). The filled 3d orbital cannot participate in bonding. For manganese, both the 3d and 4s subshells are available for occupation by bonding electrons , so a wide range of oxidation states is observed.

4

The de Broglie wavelength associated with matter is found from λ = h / m v, where m is mass (kg) and v is velocity (m s-1).

(1) For the bullet, m = 0.0070 kg and v = 1.1 x 103 m s-1.

Using the de Broglie wavelength equation,

λ = (6.626 x 10-34 J s) / (0.0070 kg) x (1.1 x 103 m s-1) = 8.6 x 10-35 m

(2) For the electron, m = 9.11 x 10-31 kg and v = 1.1 x 103 m s-1.

λ = (6.626 x 10-34 J s) / (9.11 x 10-31 kg) x (1.1 x 103 m s-1)

= 6.6 x 10-7 m.

The wave character of the electron, being in the range of nanometres, is apparent since it can be easily detected by say, diffraction through a grating. On the other hand, observing a wavelength of the order of 10-35 m is not feasable.

5

The first step is to calculate the work function of the metal from the threshold wavelength quoted in the question.

E of 520 nm photon = h c / λ

= (6.626 x 10-34 J s) x (3.00 x 108 m s-1) / (520 x 10-9 m)

= 3.82 x 10-19 J per photon. This is the work function of the metal (the energy required just to ionise one electron)

E of 360 nm photon = h c / λ

= (6.626 x 10-34 J s) x (3.00 x 108 m s-1) / (360 x 10-9 m)

= 5.52 x 10-19 J per photon.

The kinetic energy of the ejected electron is equal to the energy of the incoming 360 nm photon minus the work function.

The kinetic energy of the ejected electron

= 5.52 x 10-19 J - 3.82 x 10-19 J

= 1.70 x 10-19 J.

Now the kinetic energy of the electron will equal m v2 (where m is mass in kg, v is velocity in m s-1).

Solving for v we have v = (2E / m)

= (2.00 x 1.70 x 10-19 J / 9.11 x 10-31 kg)

= 6.1 x 105 m s-1.

6

E = 5.0 x 10-16 J

(1) λ = h c / E

= (6.626 x 10-34 J s) x (3.00 x 108 m s-1) / (5.0 x 10-16 J)

= 4.0 x 10-10 m = 0.40 nm.

(2) The wavenumber is the number of wavelengths in 1 cm (this is evident from the units cm-1). The wavelength is 4.0 x 10-10 m, so the number of wavelengths in 1 cm = (0.01 m) / (4.0 x 10-10 m) = 2.5 x 107 cm-1.

(3) n = E / h

= (5.0 x 10-16 J) / (6.626 x 10-34 J s) = 7.5 x 1017 Hz.

7

E per mol = NA h c / λ

E per mol (165 nm)= (6.022 x 1023 mol-1) x (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (165 x 10-9 m)

= 725 kJ mol-1. This is in the ultraviolet region.

E per mol (254 nm) = (6.022 x 1023 mol-1) x (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (254 x 10-9 m)

= 471 kJ mol-1. This is in the ultraviolet region.

E per mol (365 nm) = (6.022 x 1023 mol-1) x (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (365 x 10-9 m)

= 328 kJ mol-1. This is just within the ultraviolet region.

E per mol (546 nm) = (6.022 x 1023 mol-1) x (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (546 x 10-9 m)

= 219 kJ mol-1. This is in the visible region (green - yellow).

E per mol (735 nm) = (6.022 x 1023 mol-1) x (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (735 x 10-9 m)

= 163 kJ mol-1. This is in the visible region (red).

8

The transition of an electron from a higher to lower orbit is accompanied by the emission of a photon. The larger the energy difference of the orbits, the higher the energy of the emitted photon.

In the Bohr model of the atom the orbits are designated by the symbol n, where n = 1 is closest to the nucleus, n = 2 the next out, etc.

(1) The Lyman series is the series of transitions from all orbits down to n = 1. The first three lines arise from the n = 2, 3 and 4 to n = 1 transitions.

The energy of a transition is found from the energy difference between the two orbits. The energy of a given orbit n = E(n)

= -me e4 Z2 / 2 (h / 2 p)2 n2 = (-2.18 x 10-18 J) x (Z2 / n2),

where Z = charge of the nucleus and n is the orbit number.

For the hydrogen atom Z = 1, so we have

E(n) = (-2.18 x 10-18 J) x (1 / n2).

To find the difference in energy between two orbits (which will equal the energy of the emitted photon), Δ E = Efinal - Einitial

= (-2.18 x 10-18 J) x [(1 / nfinal2) - (1 / ninitial2)].

Once the energy is known, the wavelength and wavenumber can be calculated as usual.

So, for the n = 2 → 1 transition (which corresponds to the first line in the Lyman series), nfinal = 1, and ninitial = 2, so

E(n = 2 → 1) = (-2.18 x 10-18 J) x [(1 / 12) - (1 / 22)]

= -1.64 x 10-18 J. Note that the negative energy value comes from defining the energy of an electron to be zero when completely removed from the atom, so an electron in an orbit has a negative energy by this definition. However, in the context of subsequent calculations this energy value is often taken as positive to provide physically relevant results (this is obvious when looking at the units being calculated - for instance, a wavelength calculated using a negative energy value would have a negative length!)

λ = h c / E

= (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (1.64 x 10-18 J) = 121 nm.

wavenumber = 1 cm / λ = 0.01 m / 121 x 10-9 m = 8.25 x 104 cm-1.

For the n = 3 → 1 transition (which corresponds to the second line in the Lyman series), nfinal = 1, and ninitial = 3,

so E(n = 3 → 1) = (-2.18 x 10-18 J) x ((1 / 12) - (1 / 32)) = -1.94 x 10-18 J.

λ = h c / E

= (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (1.94 x 10-18 J) = 102 nm.

wavenumber = 1 cm / λ = 0.01 m / 102 x 10-9 m = 9.80 x 104 cm-1.

For the n = 4 → 1 transition (which corresponds to the second line in the Lyman series), nfinal = 1, and ninitial = 4,

so E(n = 4 → 1) = (-2.18 x 10-18 J) x ((1 / 12) - (1 / 42)) = -2.04 x 10-18 J.

λ = h c / E

= (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (2.04 x 10-18 J) = 97.4 nm.

wavenumber = 1 cm / λ = 0.01 m / 97.4 x 10-9 m = 1.03 x 105 cm-1.

(2) The limiting wavelength will correspond to the maximum transition energy. This will be the transition from n = ∞ → 1, ie an electron which is completely removed from the nucleus falling into the n = 1 orbit. Looking at the equation for transition energy:

Δ E = Efinal - Einitial = (-2.18 x 10-18 J) x ((1 / nfinal2) - (1 / ninitial2))

In this limiting case where ninitial = ∞ , the (1 / ninitial2) term disappears and nfinal = 1, giving Δ E = (-2.18 x 10-18 J) x ((1 / nfinal2)

= (-2.18 x 10-18 J) x ((1 / 12) = -2.18 x 10-18 J.

The limiting wavelength is then found using this energy value:

λ = h c / E

= (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (2.18 x 10-18 J) = 91.2 nm.

(3) The energy of the n = 1 state is defined with respect to a completely removed electron, and will therefore correspond to the energy of the transition found in part (2.), E = 2.18 x 10-18 J.

(4) The ionisation potential is the energy required to completely remove an electron from an orbit. In the case of a neutral hydrogen atom there is one electron in the n = 1 orbital. Removing this electron (n = 1 → ∞ ) is the exact reverse of the transition described in part (2.), (n = ∞ → 1). As a result the energy will be the same (2.18 x 10-18 J), except that ionisation involves photon absorption rather than emission. Ionisation potentials are often meausred in electron volts (eV) (1 eV = 1.6 x 10-19 J). Therefore the ionisation potential in eV = (2.18 x 10-18 J) / (1.6 x 10-19 eV J-1) = 13.6 eV.

9

The de Broglie wavelength is found from λ = h / m v, where m is mass (kg) and v is velocity (m s-1)

(1.) me = 9.11 x 10-31 kg, v = 1.2 x 109 cm s-1 = 1.2 x 107 m s-1.

λ = h / m v = (6.626 x 10-34 J s) / [(9.11 x 10-31 kg) (1.2 x 107 m s-1)]

= 6.1 x 10-11 m.

(2.) m = 0.170 kg, v = 3.58 x 103 cm s-1 = 3.58 x 101 m s-1.

λ = h / m v = (6.626 x 10-34 J s) / [(0.170 kg) (3.58 x 101 m s-1)]

= 1.09 x 10-34 m.

10

E per mol = NA h c / λ

E per mol (590 nm) = (6.022 x 1023 mol-1) x (6.626 x 10-34 J s) x (3.00 x 108 ms-1) / (590 x 10-9 m)

= 203 kJ mol-1.