Download Example 4: A one-electron atom is irradiated with visible light. The

CHAPTER THREE: ATOMIC SPECTROSCOPY AND THE HYDROGEN
ATOM
The chemistry of the elements depends upon the number of electrons the atoms contain and the way those electrons
are arranged in the atoms. Thus if we know the electronic configuration of each element, that will go a long way to
helping us predict and understand their chemistry.
The location of electrons in an atom is determined by
spectroscopy which can be described as the interaction of electromagnetic radiation with matter (specifically, in our
case, with atoms).
3.1
The Wave Properties of Light
Light can be considered to come in waves which can be characterized by four different properties.
Wavelength
Amplitude
λ
distance
λ
=
wavelength of light =
distance between two identical points on the wave
The wavelength of light can be expressed in the following units (depending on where we are in the electromagnetic
spectrum): m, m, nm, Å, pm.
Name
Symbol
Metres
Metre
m
1
Micrometre
µm
1 × 10-6
Nanometre
nm
1 × 10-9
Angstrom
Å
1 × 10-10
Picometre
pm
1 × 10-12
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Question 3.1
Convert 155 nm into: (i) µm (ii) Å and (iii) pm
Frequency
ν
=
frequency of the light = number of waves which pass a given spot in one second.
The frequency of light is measured in reciprocal seconds, s-1, which are equal to Hertz (Hz).
Wavenumber
ν̅ = wavenumber of the light = number of waves per centimetre
ν̅ =
1
𝜆
The units of the wavenumber of light are reciprocal centimetres, cm-1.
Speed of Light
The product of the wavelength and the frequency gives the speed of light (c =
νλ ), which has a constant value (in a
vacuum) of 2.998 × 108 m s-1.
Example 1: The wavelength of a particular light source has been measured as 300 nm. Calculate the frequency and
the wavenumber of the light.
Solution:
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The full electromagnetic spectrum is shown below in terms of wavelength.
Radio waves
Microwaves
X-Rays
Gamma rays
-12
10
10
-11
-10
10
-9
10
-8
10
Wavelength(m)
1 nanometre
VISIBLE
Infra-red
Ultra-violet
-3
-7
10
10
1 micrometre
10
-2
1 millimetre
-1
0
10
10
1
10
1 metre
2
10
3
10
4
10
1 kilometre
The parts of the spectrum that will give us important information on the position of the electrons in an atom are the
UV/visible/infra-red regions.
3.2
The Particle Properties of Light
Light can also be considered to come in discrete “packets” of energy called photons. A discrete property is a
particle-like property. You can have, for instance, 105 particles, but not 105 ½ particles. Particle-like properties are
quite different from wave-like properties.
 A wave is a periodically repeated pattern in space with no particular emphasis on any crest or trough. It does not
define a location sharply.
 A particle-like property is just the opposite; it does define an object’s position exactly.
The object in this case is a photon. One photon is the smallest amount of light you can get (just like an electron is
the smallest amount of charge one can get). Thus light possesses both wave-like and particle-like properties.
Light has a dual nature
Photons are quanta of energy (a quantum is a fixed-size packet). The energy of a photon is related to its frequency,
ν, through a constant of proportionality, Planck’s constant, in Planck’s Equation:
Energy of a photon = Planck’s constant × frequency, i.e., E = hν
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Planck’s constant, h, is given as 6.626 × 10-34 J s. Thus, when using the above equation, the unit of energy is Joules
(J). Thus if we know the frequency of a photon (or its wavelength or its wavenumber, as these are interchangeable)
we can calculate the energy the photon carries.
Example 2: Which of the following photons has the highest energy? (i) A photon with a frequency of 1.00 × 1015
s–1, (ii) A photon with a wavelength of 502 nm or (iii) A photon with a wavenumber of 25 400 cm–1.
Solution:
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Question 3.2
Calculate the energy (J) of a photon, (i) which has a wavelength of 964 nm, (ii) which has a frequency of
1.32 × 1013 Hertz and (iii) which has a wavenumber of 4 050 cm-1.
3.3
Quantization
If a property is said to be “quantised” it means that it can adopt only certain values. For instance the number of peas
in a pod are quantised as you can only have a whole number of peas. The values such as 2.2 or 5.7 peas are said to
be forbidden as you cannot have a fraction of a pea growing in a pod.
Is the energy that an electron can have in an atom quantised? The answer to the question is yes, the energies of an
electron in an atom are quantised. Electrons can only have certain energies. These energies are the allowed
energies.
The energy that electrons can adopt is therefore said to be quantised. The diagram overleaf illustrates the difference
between non-quantised energy (continuous energy) and quantised energy.
Energy continuum
Quantized energy
E
The energy continuum on the left is that which arises from any of Newton’s Laws of Motion, which are covered in
physics. A car travelling on a straight road can adopt any energy depending upon its mass and its velocity.
At the atomic level of matter, however, the energy that electrons can adopt is limited to certain values (in the case
above, the electron could adopt only five discrete values).
3.4
The Interaction of Electrons and Electromagnetic Radiation
There is a fundamental rule of science that we must remind ourselves of at this point.
Energy can neither be created nor destroyed. Energy is conserved.
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Therefore, when any interaction between atoms and electromagnetic radiation occurs, the energy that goes into this
interaction must be the same as the resulting energy after the interaction occurs. This is best explained by the
equation below:
Atom
+
Photon
Atom*
If an atom and a photon interact, the atom is said to have absorbed the photon. When the atom does this, its energy
must be increased by the same amount of energy as that carried by the photon. The atom is now in a higher energy
state as is indicated by the asterisk (*) next to it. We say the atom is excited.
Let’s take the simplest case. We have an atom which has one electron. That electron has only two allowed
energies, a lower energy, E1, and a higher energy, E2.
E
E2
E1
Chemists use the symbol
↿ for an electron.
Now given a free choice an electron will always adopt the lowest
energy it can. That state of lowest energy is called the ground state.
If we now interact our atom with a photon, the atom can absorb the energy of the photon by promoting the electron
in E1 up to E2.
E
E
E2
↿
E2
photon
↿
E1
E1
Ground state
Excited state
When an electron goes from one allowed energy to another, this is called a transition.
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When the electron is in E2 the atom is now no longer in its lowest possible energy state. It is said to be in an excited
state. The energy of the excited state of the atom must be equal to the energy of the ground state of the atom plus
the energy of the photon.
Energy of the excited atom = Energy of the atom in the ground state + energy of the photon
Eatom* = Eatom + energy of the photon
We know that the energy of the photon is
E = hν
thus
Eatom* = Eatom + hν
when we re-arrange this equation
Eatom* - Eatom = hν
We can use the symbol E2 to represent the atom in its excited state and E1 to represent the atom in its ground state
E2 – E1 = hν
We can write the difference between E2 and E1 as ΔE12. Therefore,
ΔE12 = hν
To promote an electron from E1 to E2 we must add an amount of energy equal to
hν .
In this case, as there is a
definite value of ΔE12, the atom can only absorb one definite frequency, (ν), of photons.
The energy of the photon must equal exactly ΔE12 to be absorbed by our atom. If the energy of the photon is
anything but ΔE12, it will not be absorbed because the resultant excited atom would not have an allowed energy.
Remember a photon is the smallest amount of light one can get. It cannot be divided into smaller parts. Thus if we
interact a photon of higher energy than with the atom then the following would have to occur:
↿
E2
E΄ The dashed line represents an
energy state that is not allowed
E2
hν'
↿
E1
E1
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The energy of the photon is greater than ΔE12. This means that if the atom were to absorb the energy hν the
electron would be promoted to E. But E is not an allowed energy. Therefore the atom will not absorb this photon.
Similarly if we interact our atom with a photon of lower energy than ΔE12 then
E2
↿
hν˝
↿
E2
E˝
The dashed line represents an
E1 energy state that is not allowed
E1
Will the atom absorb a photon with energy, hν˝? The electron is not allowed to have energy E. Thus again this
photon will not be absorbed. In fact our atom can only absorb one particular type of photon; those photons with the
exact energy corresponding to ΔE12. This is known as selective absorption and this phenomenon is the basis for
atomic spectroscopy.
3.5
Absorption Spectroscopy
Atomic absorption spectroscopy uses an instrument known as a spectrometer to direct photons of light through a
sample of atoms. The atoms selectively absorb some of the photons; that is, those that correspond to the differences
in the allowed energies of the atom. A detector measures indirectly those energies that have been absorbed. The
detector produces an absorption spectrum, which is a plot of absorbance, A vs the energy of the photons that have
been directed at the sample. The absorbance is a measure of the amount of light of a particular energy absorbed by
the atom.
A
Energy
For our imaginary atom that contained one electron and two allowed energies what would the absorption spectrum
look like? We know that our atom can only absorb one particular energy of photons that corresponds to E12. Thus
the absorption spectrum would look like this:
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A
ΔE12
Energy
The spectrum shows us that one line would be seen at the energy corresponding to ΔE12.
Atomic absorption spectra are termed line spectra as only specific energies are absorbed due to selective absorption.
Note that absorption spectra are often plotted as absorbance vs. wavenumber or absorbance vs. wavelength.
Example 3: The absorption spectrum of our imaginary atom was measured to give this spectrum.
A
32 500
-1
Wavenumber (cm )
What is the difference in energy between E1 and E2 in joules?
Solution:
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Real atoms have many more allowed energy levels than the two we have been dealing with up to now. However the
same rules apply. Electrons will always occupy the lowest energy and the atom will normally exist in its ground
state. Photons are generally only absorbed by the ground state of the atom, as most atoms in a sample will be in the
ground state (at normal room temperatures more than 99.99% of the atoms are in the ground state).
Let’s say we now have an atom with one electron and four allowed energy states
E4
E
E3
E2
E1
How many lines will we see in the absorption spectrum?
Three; at the following energies, ΔE12, ΔE13 and ΔE14
3.6
Emission Spectroscopy
If we supply a lot of energy to atoms, say through heating them up from normal room temperatures, then the
electrons will be promoted from the ground state energy to allowed energy states. Once at those higher excited
energies the electron’s only wish is to return from whence it came, i.e. back to the ground state. We now have the
following situation
Atom*
Atom
How does an atom lose energy? Simple; it emits a photon. Not any photon but a photon that leaves the atom in an
allowed energy state. Therefore
Atom*
High Energy
Atom
+
hν
Lower Energy
It is important to note here that the electron does not have to return directly to the ground state. Instead it can return
stepwise down the energy ladder. Thus if we have the situation:
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E4
E
E3
E2
E1
How many transitions can the electron undergo? Remember the only rule is that it must end up at an allowed
energy.
The answer to the above problem is that the electron can undergo six different transitions thus
E4
E
1
E3
2
3
4
E2
5
6
E1
Eventually all the electrons will return to the ground state of the atom. To measure an emission spectrum, the
detector of our spectrometer measures the energies of the photons emitted when the atoms are excited (usually by
heating them up). An emission spectrum is a series of lines, which corresponds to the lowering of the energies of
the electrons as they move from one energy level to a lower energy level.
In the above case the emission spectrum would have six lines, as six different transitions can occur at six different
energies.
Intensity of Emission
For instance, it might look like this
ΔE34
ΔE23
ΔE24
ΔE12
Energy
38
ΔE13
ΔE14
If you look at the figure on the previous page, you will notice that the line labelled ∆E 34 corresponds to the transition
from E4 to E3. Therefore the line labelled ∆E14 will correspond to the transition from E4 to E1.
As in the absorption spectrum we will see a line spectrum. The lines will occur at exactly the same energies as the
differences between the allowed energies involved in the transition.
Both absorption and emission spectroscopy give us a means to measure the differences between the allowed
energies of the electron.
They don’t tell us anything about the absolute electronic energies that are allowed in the atom.
3.7
The Allowed Electronic Energies
We have already noted that the energies of electrons in atoms are quantised – only certain energies are allowed.
Spectroscopic measurements such as atomic absorption and emission prove that this is the case. We would not
obtain line spectra unless selective absorption (or emission) was occurring. We would not get selective absorption
(or emission) if we did not have discrete energies allowed for the electrons. In the next section we will use the
simplest atom, the hydrogen atom, to build up a picture of the electronic structure of atoms.
Firstly, however, we must recognise a very important concept. The allowed energies that electrons can adopt are all
negative energies.
Electrons are negatively charged and the nucleus is positively charged. If the electron is at an infinite distance from
the nucleus then, by convention, its potential energy is defined as zero. As we bring the electron closer to the
nucleus then a force of attraction between the two will develop. If we now want to remove the electron back to an
infinite distance away from the nucleus we must supply energy to overcome the force of attraction between the
nucleus and the electron. If we supply energy to a system, this, again by convention, is given a positive sign. As the
electron has zero potential energy, once it regain an infinite distance from the nucleus it must have had a negative
energy when attracted to the nucleus (to satisfy the conservation of energy rule). The closer the electron comes to
the nucleus the greater will be the force of attraction and the more negative the electron’s energy will be.
The ground state of the electron will therefore be the most negative energy available.
A one electron atom has three allowed energies, (i) - 100 000 cm-1, (ii) - 50 000 cm-1 and (iii) - 20 000 cm-1. Which
is the ground state of the atom?
Answer
The ground state will be (a) because –100 000 cm-1 is the most negative energy and therefore the lowest energy.
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Question 3.3: From above example, predict the number of lines and the energies (cm-1) occurring in (i) the
absorption spectrum and (ii) the emission spectrum of this atom.
Example 4: A one-electron atom is irradiated with visible light. The complete atomic emission spectrum shows
peaks at 323 nm, 600 nm and 700 nm. Show that this data is fully consistent with the atom having only three
allowed energy states.
Solution:
40
Question 3.4: The emission spectrum of a one-electron atom showed 6 lines: 90000 cm-1, 75000 cm-1, 50000 cm-1,
40000 cm-1, 25000 cm-1 and 15000 cm-1. How many lines would you expect to see in the absorption spectrum and at
what energies? Express your answer in cm-1.
Question 2.5: An atom has three allowed energy states: - 100 000 cm-1, - 50 000 cm-1 and -30 000 cm-1.
(i) Draw an energy level diagram for the atom
(ii) Indicate on your diagram, which transitions can occur during absorption and during emission.
(iii) Sketch the absorption spectrum and the emission spectrum one would measure for such an atom showing clearly
the energies (cm-1) the lines occur at.
3.8
The Hydrogen Atom
Up to now we have been talking about the spectroscopy of theoretical atoms. It is now time to apply the general
principles of spectroscopy to real systems. The obvious place to start is the hydrogen atom. Why obvious?
Hydrogen is unique in the periodic table, it is the only element with only one electron.
Orbital Energies
When a property is quantized then the values it can adopt are limited. Every quantized property has what is known
as a quantum number. In the case of the hydrogen (H) atom the quantum number which limits the allowed energies
that the electron can adopt is given the symbol n, and is called the principal quantum number. We will now call
the allowed energies that an electron can adopt orbitals. Electrons can only occur in an orbital within an atom and
each orbital has an allowed energy. The energies of the orbitals within the H atom are given by the following
equation.
En =
−me e4
1
× 2 Joules
8h2 ε2v
n
where En = energy of the orbital described by the quantum number n (Joules), me = mass of the electron, e = charge
of the electron, h = Planck’s constant and εv = permittivity of a vacuum
You will probably not have met permittivity (εv) before. Permittivity is simply a measure of how easy it is for
electromagnetic radiation to pass through free space (no further understanding is required for this course).
You will be pleased to know that you don’t need to memorize this equation. We need this equation to derive
another which is important and must be memorized. Now, recall
E (J) = h (J s) × c (cm s-1)  × ν̅ (cm-1)
Therefore to change joules into wavenumbers, re-arrange the equation so that 𝜈̅ is the subject of the formula
νn =
E
−me e4
1
= [ 3 2 ] × 2 cm−1
hc
8h εv
n
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As all the terms in the bracket are physical constants we can group them into a single constant, RH, the Rydberg
constant for the H atom.
me e4
cm−1
8h3 ε2v c
RH =
Now we have the more easily remembered equation:
𝛎𝐧 =
−𝐑 𝐇
𝐜𝐦−𝟏
𝐧𝟐
The Rydberg constant for the hydrogen atom, RH, is given in the data sheet as 109 677 cm-1.
To calculate the allowed electronic energies in the hydrogen atom we simply need to know the value of the principal
quantum number, n. Remember it is the quantum number that limits the allowed values of the energies and
therefore n must have limited values.
n is limited to any positive whole number, n = 1, 2, 3, 4, 5, ………….∞
Therefore there are a lot of allowed energies in the H atom! Let’s work out the first few energies and the last one.
n
𝜈(cm-1)
1
- RH
2
-RH/4
3
-RH/9
4
-RH/16
5
-RH/25
∞
0
When n = ∞, the electron’s energy is zero. This is the same as saying that the electron is an infinite distance from
the nucleus or that the electron has been removed from the atom. When an electron is removed from the atom, this
is known as ionisation.
We can derive a general equation to calculate the energy gap between ANY two energy levels of the H atom.
n΄
∆ν
ν
n˝
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If we have the general case where n' is greater than n'' then
∆ν =
−R H −R H
−R H R H
1
1
−
=
+ 2 = RH ( 2 − 2)
2
2
2
n΄
n˝
n΄
n˝
n˝
n΄
This equation is the Rydberg equation for the H atom and from it we can predict the position of the lines in the
absorption spectrum and the emission spectrum. You need to know this equation.
Example 5: Calculate the energy (cm-1) of the electronic transition between n = 8 and n = 4 in the emission spectrum
of H. To which region of the electromagnetic spectrum does this energy belong?
Solution:
When looking at the emission spectrum of hydrogen more than one series of lines becomes apparent. These
different series arise from the different values of n'' in the Rydberg equation. These series are named after their
discoverers and are summarized below.
Series
n˝
n΄
Region of the electromagnetic spectrum
Lyman
1
2, 3, 4,….
UV
Balmer
2
3, 4, 5, …
visible
Paschen
3
4, 5, 6,….
ir
Brackett
4
5, 6, 7, ….
ir
Pfund
5
6, 7, 8,…..
ir
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You don’t need to remember these names but do get some idea of which region of the electromagnetic spectrum
they fall in. Remember the gaps between the levels get smaller as n increases and therefore the energies of the
transitions decrease.
Example 6: Calculate the energy (cm-1) of the 3rd line of the Balmer series.
Solution:
Question 2.6: Calculate the energy (cm-1) of the (i) 5th line of the Lyman series, (ii) the 5th line of the Paschen series
and (iii) the transition from the first energy level to the third energy level in a hydrogen atom.
3.9
Quantum numbers
We’ve already seen that the energies of the atomic orbitals of hydrogen are quantized by the principal quantum
number, n.
Other properties of the atomic orbitals are also quantized, viz., the shapes of the orbitals and the direction the
orbitals are pointing in space.
3.9.1
Angular Momentum
Electrons are particles with a mass and a velocity and their trajectory takes them around the nucleus. The electrons
therefore have a property known as angular momentum. The values of angular momentum that the electrons can
possess are also quantized. The values are limited by the angular momentum quantum number, l. The effect of all
this is that the shapes of the orbitals are limited.
The allowed values of l depend on the principal quantum number, n.
l = 0, 1, 2, 3 , ……… ( n –1)
The maximum value l can have is thus (n – 1).
Therefore if n = 3 then l can take the values of 0, 1, or 2. All these values are allowed when n = 3.
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Note:
The smallest allowed value of l is 0 and the smallest allowed value of n is 1
Each different value of l leads to a different value of angular momentum, and a different shape of orbital.
l
orbital
0
s
1
p
2
d
3
f
4
g
Thus if l = 2 we have a d-orbital.
3.9.2
Orientation of Orbitals
As well as having a fixed shape, the orbitals must point in a set direction in space. We say the orbital has a fixed
orientation in space. The directions are limited by the magnetic quantum number, m l. The values of ml depend on
the value of l, the angular momentum quantum number.
ml = 0, ± 1, ± 2, ± 3, ………..± l
For a specific value of l, there are (2 l + 1) integral values of ml.
The possible values for ml are:
- l , (-l + 1), …….., 0,……. , (+l –1), + l
So, if l = 1, the possible values of ml are
- l = -1, (-l + 1) = 0, (+l – 1) = 0, + l = +1
Thus ml = -1 or ml = 0 or ml = +1
We usually write this as: ml = -1, 0, +1
If l = 2, the possible values for ml are:
(- l) = -2, (-l + 1) = -1, (-l +2) = 0, (+l – 1) = +1, (+l) = +2
 ml = -2, -1, 0, +1, +2
Thus if l is 0 then ml = 0
l is 1 then ml = +1, 0, -1
l is 2 then ml = +2, +1, 0, -1, -2
l is 3 then ml = +3, +2, +1, 0, -1, -2, -3
Each value of ml represents a different atomic orbital and any given orbital can have only one value of ml.
Thus if l is 0 then ml = 0 and we have one s-orbital
l is 1 then ml = +1 or 0 or -1 and we have three p-orbitals
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l is 2 then ml = +2 or +1 or 0 or –1 or -2 and we have five different d-orbitals
l is 3 then ml = +3 or +2 or +1 or 0 or –1 or –2 or -3 and we have seven f-orbitals
Question 3.7: If the principal quantum number, n, for an atomic orbital is 4, (i) what are the possible values of l ?,
(ii) for each value of l what are the possible values of ml ? and (iii) how many different atomic orbitals are possible
when n = 4?
3.10
The Shapes of the Orbitals
s-orbitals are spherical
z
y
x
p-orbitals are shaped like a dumb-bell with each of the three p-orbitals pointing in a different direction. We call
these three p-orbitals px, py and pz.
z
y
z
z
y
x
y
x
x
px
py
pz
The d-orbitals have a more complex shape but again each of the d-orbitals points in a different direction.
x
y
y
dxy
z
dyz
x
x
z
dzx
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z
y
dx2-y2
dz2
The f-orbitals are even more complex (we won’t even attempt to show them here) but again all seven orbitals have
different orientations in space.
3.11
The Orbital Diagram
We can now build up the complete picture of the atomic orbitals. Firstly we divide them into shells. Each shell has
a different principal quantum number.
When n = 1, we call it the K shell, l can only equal 0 and the only allowed value of ml is also 0.
When n = 2 we call it the L shell, l = 0 or 1
l=0
then
ml = 0 have
1 × 2s
l=1
then
ml = +1, 0, -1 have
3 × 2p
Since l = 0 or 1, we say there are two subshells, the s-subshell and the p-subshell.
When n = 3 we call it the M shell, l can have values of 0 or 1 or 2
l=0
ml = 0
1 × 3s
l=1
ml = +1 or 0 or -1
3 × 3p
l=2
ml = +2 or +1 or 0 or –1 or -2
5 × 3d
Since l = 0 or 1 or 2, there are three subshells, the s, p and d-subshells.
When n = 4 we call it the N shell, l = 0 or 1 or 2 or 3
l=0
ml = 0
1 × 4s
l=1
ml = +1 or 0 or -1
3 × 4p
l=2
ml = +2 or +1 or 0 or –1 or -2
5 × 4d
l =3
ml = +3 or +2 or +1 or 0 or –1 or –2 or -3
7 × 4f
Since l = 0 or 1 or 2 or 3, there are four subshells, the s, p, d and f-subshells.
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Question 3.8: Write down all the orbitals present in the n = 5 shell (Hint: you should get n 2 orbitals which equals
25).
Question 3.9: Use the Rydberg equation for atomic hydrogen to calculate the wavelength (nm) for the transition
from n = 7 to n = 2.
Question 3.10: Write the notation (in the form 3d for example) and the number of orbitals having the following
quantum numbers in an atom: (i) n = 4, l = 2, (ii) n = 6, l = 3, (iii) n = 2, l = 1 and (iv) n = 0, l = 0
Question 3.11: Identify the values of the principal quantum number and the angular momentum quantum number for
the following orbitals. Also suggest a value for the magnetic quantum number, (i) 5f, (ii) 3p, (iii) 2s and (iv) 4d
Question 3.12: How many orbitals are present in a shell with n = 7?
Question 3.13: Calculate the ionisation energy (cm-1) for a hydrogen atom initially in its ground state.
[Hint: remember that to ionise the atom the electron must be removed completely from the atom.]
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