For online help with this topic,
see the chemcal module "Atomic
and Nuclear Structure" which deals with the topics: Fundamental
concepts; nuclear equations; electromagnetic radiation; waveparticle
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
^{34}X (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 ^{40}Ca
and ^{44}Ca.
For both ^{40}Ca
and ^{44}Ca^{2+} 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 10^{3} 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 / n^{2}), 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 10^{9} cm
s^{1}
(2)
a golf ball of mass 170 g, travelling with a speed of 3.58 x 10^{3}
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 1s^{2} 2s^{2} 2p^{6} 3s^{2}
3p^{4}.
(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 S^{2} ion (II oxidation state), eg reaction with strontium:
Sr(s) + S(s) → SrS(s)
Sulfur can also react with nonmetals to produce compounds in +IV
and +VI oxidation states, eg burning in oxygen:
S(s) + O_{2}(g) → SO_{2}(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 ^{44}Ca^{2+} 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 ^{40}Ca
this is 40. Note that the charge present on the ^{44}Ca^{2+}
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 ^{40}Ca is 20. The ^{44}Ca^{2+}
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 ^{40}Ca this equals 40  20 = 20 neutrons.
For ^{44}Ca^{2+} 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)
^{40}Ca : 1s^{2} 2s^{2} 2p^{6} 3s^{2}
3p^{6} 4s^{2}
^{44}Ca^{2+} : 1s^{2} 2s^{2} 2p^{6}
3s^{2} 3p^{6}
Note the loss of the two outer electrons in the Ca^{2+}
ion.


3 
(1) (a)
Mn : 1s^{2} 2s^{2} 2p^{6}
3s^{2} 3p^{6} 3d^{5} 4s^{2}
(b)
Zn : 1s^{2} 2s^{2} 2p^{6} 3s^{2}
3p^{6} 3d^{10} 4s^{2}
(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 10^{3} m s^{1}.
Using the de Broglie wavelength equation,
λ = (6.626 x
10^{34} J s) / (0.0070 kg) x (1.1 x 10^{3} 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 10^{3}
m s^{1}.
λ = (6.626 x 10^{34} J
s) / (9.11 x 10^{31} kg) x (1.1 x 10^{3} 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 10^{8} 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 10^{8} 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 ½ m v^{2} (where m is mass in
kg, v is velocity in m s^{1}).
Solving for v we have
v = (2E / m)^{½}
= (2.00 x 1.70 x 10^{19}
J / 9.11 x 10^{31} kg)^{½}
= 6.1 x 10^{5}
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 10^{8}
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 10^{7} cm^{1}.
(3)
n = E / h
= (5.0 x 10^{16} J) / (6.626 x 10^{34}
J s) = 7.5 x 10^{17} Hz.


7 
E per mol = N_{A}
h c / λ
E per mol (165 nm)=
(6.022 x 10^{23} mol^{1}) x (6.626 x 10^{34}
J s) x (3.00 x 10^{8} 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 10^{23} mol^{1}) x (6.626 x 10^{34}
J s) x (3.00 x 10^{8} 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 10^{23} mol^{1}) x (6.626 x 10^{34}
J s) x (3.00 x 10^{8} 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 10^{23} mol^{1}) x (6.626 x 10^{34}
J s) x (3.00 x 10^{8} 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 10^{23} mol^{1}) x (6.626 x 10^{34}
J s) x (3.00 x 10^{8} 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)
= m_{e} e^{4} Z^{2} / 2 (h / 2
p)^{2} n^{2}
= (2.18 x 10^{18} J) x (Z^{2} / n^{2}),
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 / n^{2}).
To find the difference in energy between two orbits (which will
equal the energy of the emitted photon), Δ
E = E_{final}  E_{initial}
= (2.18 x 10^{18} J) x [(1 / n_{final}^{2})
 (1 / n_{initial}^{2})].
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), n_{final}
= 1, and n_{initial} = 2, so
E(n = 2 → 1) = (2.18 x
10^{18} J) x [(1 / 1^{2})  (1 / 2^{2})]
= 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 10^{8} 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 10^{4} cm^{1}.
For the n = 3 → 1 transition
(which corresponds to the second line in the Lyman series), n_{final}
= 1, and n_{initial} = 3,
so E(n = 3 → 1) = (2.18
x 10^{18} J) x ((1 / 1^{2})  (1 / 3^{2}))
= 1.94 x 10^{18} J.
λ = h c / E
= (6.626 x 10^{34} J s) x (3.00 x 10^{8} 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 10^{4} cm^{1}.
For the n = 4 → 1 transition
(which corresponds to the second line in the Lyman series), n_{final}
= 1, and n_{initial} = 4,
so E(n = 4 → 1) = (2.18
x 10^{18} J) x ((1 / 1^{2})  (1 / 4^{2}))
= 2.04 x 10^{18} J.
λ = h c / E
= (6.626 x 10^{34} J s) x (3.00 x 10^{8} 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 10^{5} 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 = E_{final}  E_{initial}
= (2.18 x 10^{18} J) x ((1 / n_{final}^{2})
 (1 / n_{initial}^{2}))
In this limiting case where n_{initial} = ∞
, the (1 / n_{initial}^{2}) term disappears
and n_{final} = 1, giving Δ
E = (2.18 x 10^{18} J) x ((1 / n_{final}^{2})
= (2.18 x 10^{18} J) x ((1 / 1^{2}) = 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 10^{8} 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.)
m_{e} = 9.11 x 10^{31} kg, v = 1.2 x 10^{9}
cm s^{1} = 1.2 x 10^{7} m s^{1}.
λ = h / m v = (6.626 x 10^{34}
J s) / [(9.11 x 10^{31} kg) (1.2 x 10^{7} m s^{1})]
= 6.1 x 10^{11} m.
(2.)
m = 0.170 kg, v = 3.58 x 10^{3} cm s^{1} = 3.58
x 10^{1} m s^{1}.
λ = h / m v = (6.626 x 10^{34}
J s) / [(0.170 kg) (3.58 x 10^{1} m s^{1})]
= 1.09 x 10^{34} m.


10 
E per mol = N_{A}
h c / λ
E per mol (590 nm) =
(6.022 x 10^{23} mol^{1}) x (6.626 x 10^{34}
J s) x (3.00 x 10^{8} ms^{1}) / (590 x 10^{9}
m)
= 203 kJ mol^{1}.


