The wave-mechanical model of the atom and orbitals
Why is the electron of an atom drawn in a chemistry book as a fuzzy cloud and not as a small ball going round the nucleus along an orbit?
Prior knowledge
- Electrons sent one at a time each hit one spot behind a double slit, but many hits build up an interference patternchapter Light and electrons as waves and particles
- The energy levels of the hydrogen atom are $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$, and the energy of the photon produced in a transition is the difference between the levelschapter Energy levels of the hydrogen atom and the Bohr model
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 5.
Think about it
On a page of a chemistry book, the electrons of a carbon atom do not go round the nucleus along neat orbits. They are drawn as blue clouds and two-lobed shapes, and the caption speaks of a region where the electron is likely to be detected. In physics, however, the energy levels of the hydrogen atom were calculated with the Bohr model, in which the electron goes round the nucleus on a circular orbit.
What does the picture of an orbital actually show, if it is not the path of the electron?
Optional Review: the electron in a double slit
Predict
Make your predictions before investigating 2 questions
Prediction 1: the boundary of an orbital
Prediction 2: how far from the nucleus?
Investigate
Measure the position of the electron again and again and see how random hits build up an orbital.
Simulation
How to use
Choose the state of the electron ($1\text{s}$, $2\text{s}$ or $2\text{p}$) and the number of measurements $N$. Each dot is the result of one position measurement. The slider chooses the threshold $p$ of the boundary: the boundary is drawn so that the electron is detected inside it with probability $p$. Dots that hit inside the boundary are blue and those that hit outside are grey. The histogram on the right shows how many hits there are at each distance from the nucleus in steps of $0.02\,\text{nm}$; the darkest bar is the most common distance and the dashed line is the expected value of the model. New measurement series draws new measurement results. When $N$ increases, new hits are added to the earlier ones.
Assumptions of the model
The model is a hydrogen atom whose electron is in the $1\text{s}$, $2\text{s}$ or $2\text{p}$ state. Each measurement is made in a new atom in the same state, and the result is drawn from the probability distribution of the state. The picture is a see-through projection: the three-dimensional measurement results and the boundary are drawn in one plane, so some of the grey dots drawn at the boundary are in reality in front of or behind it. In the $2\text{s}$ state there is a gap around the nodal sphere that does not belong to the region, so its boundary is drawn as three circles. The lobes of the $2\text{p}$ state are vertical. In a real experiment the position of the electron of a single atom is not measured as directly as this; the model illustrates the probability interpretation.
How the calculation works
The share inside the boundary is $k : N$, where $k$ is the number of blue dots. The boundary is a surface on which the probability density is the same everywhere and which encloses the probability $p$. The greatest distance $R$ of the boundary is measured from the nucleus in the direction in which the region extends farthest; in the $2\text{p}$ state it is the tip of a lobe. The most common distance is the midpoint of the tallest bar of the histogram. The lengths have been calculated using the Bohr radius $0.0529\,\text{nm}$.
Experiment 1: one measurement or many?
Test prediction 1. The state is $1\text{s}$ and the threshold of the boundary is $90\,\%$. Increase the number of measurements $N$ and, for each number, record the measurements that hit inside and outside the boundary and the share inside the boundary. Also try what New measurement series does with a small and a large $N$.
Initial state: state: $1\text{s}$ orbital, measurements $N$ = $10$, threshold of the boundary $p$ = $90\,\%$. Kept constant: state, threshold of the boundary $p$. Varied: measurements $N$.
| measurements $N$ | inside $k$ | outside | share inside $(\%)$ |
|---|
Experiment 2: how far from the nucleus is the electron detected?
Test prediction 2. There are $5000$ measurements and the threshold of the boundary is $90\,\%$. Change the state of the electron and, for each state, record the most common distance of the histogram and the greatest distance of the boundary of the orbital from the nucleus.
Initial state: state: $1\text{s}$ orbital, measurements $N$ = $5000$, threshold of the boundary $p$ = $90\,\%$. Kept constant: measurements $N$, threshold of the boundary $p$. Varied: state.
| state | most common $r$ $(\text{nm})$ | $R$ $(\text{nm})$ |
|---|
Conclusion: what does an orbital tell?
Based on your tables, write: what does a single measurement tell about the position of the electron, and what appears only after many measurements? What does the $90\,\%$ boundary of an orbital mean? How does the $1\text{s}$ orbital differ from the $2\text{s}$ and $2\text{p}$ orbitals?
Model answer
A single measurement gives one random point. With small values of $N$ the share of hits inside the boundary varied clearly, and a new measurement series gave different numbers. When there were $1000$ or $5000$ measurements, the share settled close to $90\,\%$. The boundary is thus a probability boundary: about a tenth of the hits remain outside it. According to the values of the model, the most common distance of a $1\text{s}$ electron from the nucleus was about $0.05\,\text{nm}$, the radius of the Bohr orbit, and the boundary of the $1\text{s}$ orbital was $0.14\,\text{nm}$ away. In the $2\text{s}$ and $2\text{p}$ states the most common distance was about $0.2$–$0.3\,\text{nm}$ and the boundary extended about $0.48\,\text{nm}$ away: an electron of the shell $n = 2$ is detected on average farther from the nucleus, and its energy is greater.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
From an orbit to a wave
The Bohr model gave the energy levels of the hydrogen atom correctly (chapter Energy levels of the hydrogen atom and the Bohr model), but its picture of a small ball on a circular orbit does not stand closer examination. The model does not tell why only certain orbits are allowed, and it does not work for atoms with several electrons.
In the chapter Light and electrons as waves and particles we also saw that electrons sent one at a time form an interference pattern behind a double slit: the motion of an electron must be described by a wave. In the wave-mechanical model of the atom, the state of an electron in an atom is described by a wave that spreads around the nucleus, not by an orbit.
The allowed states are standing waves
A guitar string is fixed at both ends, so only those standing waves arise in it in which the ends of the string are nodes. There are therefore only certain ways of vibrating: one antinode, two antinodes, three antinodes and so on. The electric attraction of the nucleus binds the electron, and its wave must likewise fit around the nucleus. Lasting states arise only for certain wave shapes, and each shape has its own energy. The quantisation of energy is thus not a separate assumption as in the Bohr model, but follows from the fact that a bound wave can only be a standing wave. In a string, an extra node raises the frequency of the vibration. In the hydrogen atom the energy of a state is greater the more nodes the wave has; in an atom with many electrons this holds for states of the same shape (s, p, …).
In an atom the wave is three-dimensional, so its nodes are surfaces, and the electron is not detected on a nodal surface. The wave of the lowest state, the $1\text{s}$ state, has no node. The states of the shell $n = 2$ have one node: in the $2\text{s}$ state it is a spherical surface around the nucleus, and in the $2\text{p}$ state a plane through the nucleus. The number in the name of the state is the principal quantum number $n$, and there are $n - 1$ nodes altogether. The letter tells the shape of the wave: an s state is spherically symmetric, and a p state has two lobes on opposite sides of the nucleus.
The probability interpretation
The wave does not tell where the electron is but where it is likely to be detected. When the position of the electron is measured, the result is always one point, and a single result cannot be predicted. The square of the amplitude of the wave tells how likely the electron is to be detected in a small volume around each place. This is the probability interpretation of the wave model. The distribution becomes visible only when the measurement is repeated in many atoms in the same state. If $k$ of $N$ measurements land in some region, the probability of detecting the electron there is approximately
$$P \approx \frac{k}{N}$$
where $P$ is the probability, $k$ the number of measurements that landed in the region and $N$ the number of all measurements. The estimate becomes more accurate as $N$ grows. Between measurements the electron has no orbit that could be drawn.
An orbital is a region of probability
When the measurement results are marked as dots, an electron cloud is formed that is densest where the electron is most likely to be detected. The cloud has no sharp edge. That is why an orbital is drawn as a region inside which the electron is detected with a chosen probability, usually with a probability of $90\,\%$. The boundary is an agreed threshold, not a wall: about a tenth of the measurements land outside it. The $90\,\%$ boundary of the $1\text{s}$ orbital of the hydrogen atom is a sphere with a radius of $0.14\,\text{nm}$. The word orbital also means the state itself, for example the $2\text{p}$ orbital.
The cloud is densest at the nucleus, but the most common distance from the nucleus is not zero. A thin spherical shell farther from the nucleus holds more volume than a small sphere right around the nucleus, so a $1\text{s}$ electron is detected most often at a distance of about $0.053\,\text{nm}$. This is the same as the radius of the smallest orbit of the Bohr model: in the wave model the Bohr orbit is the most common distance, not a path travelled by the electron.
Shells, orbitals and energy levels
In the hydrogen atom the energy of a state depends only on the principal quantum number $n$, and the wave model gives the same energy levels as the Bohr model:
$$E_n = -\frac{13.6\,\text{eV}}{n^2}$$
where $E_n$ is the energy of the electron (eV) in the shell $n = 1, 2, 3, \ldots$ States with the same principal quantum number form an electron shell. The shell $n$ has $n$ subshells, s, p, d and f, which have $1$, $3$, $5$ and $7$ orbitals, so the number of orbitals in a shell is $n^2$.
The number of orbitals in a subshell depends on the shape of the wave. An s orbital is a sphere, so there can be only one. The pair of lobes of a p orbital can point in three mutually perpendicular directions ($x$, $y$ and $z$), so there are three p orbitals; the figures of this chapter show one of them, whose lobes are vertical. The d and f subshells have five and seven different orbitals, respectively. An electron also has spin, its own property with only two possible values. Each orbital can therefore hold at most two electrons, and they have opposite spins (Pauli exclusion principle). So at most
$$N_\text{max} = 2n^2$$
electrons fit in a shell, where $N_\text{max}$ is the greatest number of electrons and $n$ the principal quantum number of the shell.
| Shell $n$ | Subshells | Orbitals $n^2$ | Electrons $2n^2$ |
|---|---|---|---|
| $1$ | s | $1$ | $2$ |
| $2$ | s, p | $4$ | $8$ |
| $3$ | s, p, d | $9$ | $18$ |
| $4$ | s, p, d, f | $16$ | $32$ |
When an electron moves from a higher state to a lower one, the atom emits a photon whose energy is the difference between the energies of the states (chapter Spectra reveal the energy levels of atoms). In the hydrogen atom the states of the same shell have the same energy, so, for example, the transitions $3\text{d} \to 2\text{p}$ and $3\text{s} \to 2\text{p}$ produce the same spectral line.
Connection to the orbitals of chemistry
In an atom with many electrons the electrons repel each other, so the subshells of the same shell have different energies: $2\text{s}$ is lower than $2\text{p}$ and $3\text{s}$ lower than $3\text{p}$. In the ground state the electrons fill the orbitals starting from the lowest energy, at most two in each orbital. The filling order of light atoms is $1\text{s}$, $2\text{s}$, $2\text{p}$, $3\text{s}$, $3\text{p}$. The six electrons of carbon are placed like this: $1\text{s}^2\,2\text{s}^2\,2\text{p}^2$, where the superscript is the number of electrons in the subshell.
The electrons of the outermost shell, the valence electrons, are on average farthest from the nucleus and most weakly bound, and they decide what kind of bonds the atom forms. The periodic table and the orbital pictures of chemistry thus rest on the same wave model.
Common misconception
Many think that an orbital is an orbit along which the electron goes round, or a shell inside which the electron stays. In the wave model the electron has no orbit, and the boundary of an orbital is an agreed probability boundary: outside the $90\,\%$ boundary the electron is detected on average in every tenth measurement. Another mistake is to think that the cloud arises because the electron whizzes so fast that its orbit blurs like the blades of a rotating fan. The cloud describes the distribution of the measurement results: a single measurement always gives one point, and the position of the electron between measurements cannot be described as a path.
Summary
- In the wave-mechanical model of the atom, the state of an electron is described by a standing wave that spreads around the nucleus. Only certain wave shapes are possible, so the energy is quantised; in the hydrogen atom, the more nodes, the greater the energy.
- Probability interpretation: the wave tells how likely the electron is to be detected at each place. One measurement gives one random point, and the distribution is seen only from many measurements: $P \approx \tfrac{k}{N}$.
- An orbital is a region inside which the electron is detected with a chosen probability, for example $90\,\%$. The $90\,\%$ boundary of the $1\text{s}$ orbital of the hydrogen atom is a sphere with a radius of $0.14\,\text{nm}$.
- The shell $n$ has $n^2$ orbitals and at most $2n^2$ electrons. The energy levels of the hydrogen atom are $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$.
- Orbitals fill starting from the lowest energy ($1\text{s}$, $2\text{s}$, $2\text{p}$, $3\text{s}$, $3\text{p}$), and this is the basis of the electron configurations and the periodic table of chemistry.
Optional Return to your predictions
Compare your predictions with the values of the model in the experiments. Then return to the question at the beginning: what does the picture of an orbital in a chemistry book show, if it is not the path of the electron?
Would you change your prediction?
Apply
Worked example
Problem
The electron of a hydrogen atom moves from the $3\text{d}$ orbital to the $2\text{p}$ orbital, and the atom emits a photon. What is the wavelength of the photon?
Step 1/7
Given and asked. The initial state $3\text{d}$ belongs to the shell $n = 3$ and the final state $2\text{p}$ to the shell $n = 2$. The energy levels of the hydrogen atom are $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$, the Planck constant is $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$ and the speed of light is $c = 2.998 \cdot 10^{8}\,\text{m/s}$. The wavelength $\lambda$ of the photon is asked.
Step 2/7
Principle. Energy is conserved: the energy of the atom decreases by the difference between the energies of the states, and this energy leaves as a photon whose energy is $hf = \tfrac{hc}{\lambda}$. In the hydrogen atom the energy of a state depends only on the principal quantum number, so the letters d and p do not change the energy; they tell the shape of the wave.
Step 3/7
Formula solved.
$$\begin{aligned} \frac{hc}{\lambda} &= E_3 - E_2 \\ \lambda &= \frac{hc}{E_3 - E_2} \end{aligned}$$
Step 4/7
Unit conversions. The energy difference in electronvolts is
Extra figures are kept in the intermediate results, and rounding is done only at the end. The energy difference is converted to joules by substituting $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$:
Step 5/7
Substitution. First we calculate the numerator
and divide it by the energy difference:
$$\begin{aligned} \lambda &= \frac{hc}{E_3 - E_2} \\ &= \frac{1.98647 \cdot 10^{-25}\,\text{J}\,\text{m}}{3.0260 \cdot 10^{-19}\,\text{J}} \\ &= 6.5647 \cdot 10^{-7}\,\text{m} \\ &= 656.47\,\text{nm} \end{aligned}$$
Step 6/7
Answer. The wavelength of the photon is $656\,\text{nm}$, with three significant figures like the energy $13.6\,\text{eV}$.
Step 7/7
Reasonableness. The wavelength is in the range of visible light, and $656\,\text{nm}$ is the red line of the spectrum of hydrogen (chapter Spectra reveal the energy levels of atoms). The photon energy $1.9\,\text{eV}$ lies between the energies of photons of visible light, $1.6$–$3.3\,\text{eV}$. The transition $3\text{s} \to 2\text{p}$ would give the same line, because the energy of the initial state is the same.
Optional Find the error
Optional Apply to a new situation
The electron of a hydrogen atom has been excited to the $2\text{p}$ state, whose two lobes are above and below the nucleus. Its position is measured $6000$ times, and the $90\,\%$ boundary of the $2\text{p}$ orbital is drawn in the picture.
How many hits do you expect to land inside the boundary in the upper lobe?
Hint 1: a guiding question
What fraction of all hits lands inside the boundary?
Hint 2: what is the connection
The $2\text{p}$ orbital is symmetric: on average equally many of the hits inside the boundary land in each lobe.
Hint 3: an intermediate step
$0.90 \cdot 6000 = 5400$ hits land inside the boundary. Divide them between the two lobes.
Show the solution
Given
measurements $N = 6000$, threshold of the boundary $p = 90\,\% = 0.90$; the $2\text{p}$ orbital has two similar lobes on opposite sides of the nucleus
Asked
the expected number $k_\text{upper}$ of measurements that land inside the boundary in the upper lobe
Principle
According to the probability interpretation, in many measurements the share of those that land in a region is approximately the probability of the region, $P \approx \tfrac{k}{N}$. The probability $0.90$ remains inside the boundary, and the $2\text{p}$ state is symmetric with respect to the nodal plane through the nucleus, so the probability is divided equally between the lobes.
Formula
$k = pN$ and $k_\text{upper} = \dfrac{pN}{2}$
Conversions
Percentages to a decimal number: $90\,\% = 0.90$.
Substitution
Answer
$2700$
Hits, as an expected value; the real number varies from one measurement series to another by a few tens.
Reasonableness
Without the boundary, half of the hits, that is $3000$, would land in each half, and a tenth remains outside the boundary, so the result is slightly under $3000$. The total of the lobes, $2 \cdot 2700 = 5400$, is $90\,\%$ of all $6000$ measurements.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- Write the electron configuration of a nitrogen atom (seven electrons) in the ground state.
Answer
$1\text{s}^2\,2\text{s}^2\,2\text{p}^3$: two electrons in the $1\text{s}$ orbital, two in the $2\text{s}$ orbital and three in the $2\text{p}$ subshell, seven altogether.
- Why can a guitar string vibrate only in certain ways, and what does the same idea mean for the electron of an atom?
Answer
The ends of the string stay in place, so the nodes of a standing wave must fall on the ends, and only certain wave shapes fit the string. The wave of an electron bound by the nucleus can likewise be only in certain standing shapes, and each shape has its own energy, so the energy of the electron is quantised.
1. What does an orbital show?
2. Nodes and energy
3. Energy of a transition
The electron of a hydrogen atom moves from the $2\text{p}$ orbital to the $1\text{s}$ orbital, and the atom emits a photon.
What is the energy of the photon in electronvolts?
Hint 1: a guiding question
Which shells do the initial and final states belong to?
Hint 2: what is the connection
The energy of the photon is the difference between the energies of the states: $E = E_2 - E_1$, where $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$.
Hint 3: an intermediate step
$E_2 = -3.40\,\text{eV}$ and $E_1 = -13.6\,\text{eV}$.
Show the solution
Given
initial state $2\text{p}$, so $n = 2$; final state $1\text{s}$, so $n = 1$; energy levels of the hydrogen atom $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$
Asked
the energy $E$ of the photon in electronvolts
Principle
Energy is conserved: when the electron moves to a lower state, the energy of the atom decreases by the difference between the energies of the states, and the same energy leaves as a photon. In the hydrogen atom the energy of a state depends only on the principal quantum number, so the letters p and s have no effect.
Formula
Conversions
None are needed, because the energy levels and the asked energy are in electronvolts.
Substitution
Answer
$10.2\,\text{eV}$
Three significant figures like the energy $13.6\,\text{eV}$.
Reasonableness
The energy is positive and smaller than the binding energy of the ground state, $13.6\,\text{eV}$, because the electron is not released but moves from one shell to another. The energy of the photon is about $3$–$6$ times the energies of photons of visible light, $1.6$–$3.3\,\text{eV}$, so the radiation is ultraviolet, like the shortest lines of the spectrum of hydrogen.
Review the Worked example and the level 2 exercises.
Optional 4. Explain
Your friend says: "The electron whizzes around the $1\text{s}$ orbital so fast that its path looks like a blurry cloud." Using the wave model, explain what the picture of an orbital really shows and why its edge is agreed.
Model answer
In the wave model the electron has no path that it would travel, so the cloud is not a blurred path. The state of the electron is described by a standing wave, and the square of the amplitude of the wave tells how likely the electron is to be detected at each place. When the position is measured, the result is one random point. The cloud appears only when the measurement is repeated in many atoms in the same state and the results are marked in the same picture. The cloud thins out outwards, but it has no sharp edge. That is why the boundary of an orbital is drawn at an agreed probability, usually so that the electron is detected inside it with a probability of $90\,\%$; about every tenth measurement lands outside the boundary.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 5-1Core
In heavy elements the electrons are in shells farther and farther out. What is the greatest number of electrons that fit in the shell $n = 5$?
Answer
$50$
Show the solution
Given
principal quantum number of the shell $n = 5$
Asked
the greatest number of electrons $N_\text{max}$ that fit in the shell
Principle
The shell $n$ has $n^2$ orbitals, because its subshells s, p, d, … have $1$, $3$, $5$, … orbitals. Each orbital can hold at most two electrons with opposite spins (Pauli exclusion principle), so the greatest number is twice the number of orbitals.
Formula
$N_\text{max} = 2n^2$
Conversions
None are needed, because a count is asked.
Substitution
$N_\text{max}$$= 2 \cdot 5^2$$= 2 \cdot 25$$= 50$Answer
$50$
Electrons; the result is an exact integer, because it is calculated from counts.
Reasonableness
The result continues the series $2, 8, 18, 32$ of the table: the differences $6$, $10$ and $14$ of successive shells grow by four, so the next difference is $18$ and $32 + 18 = 50$. So more electrons fit in the shell $n = 5$ than in the shell $n = 4$, as they should.
- Exercise 5-2Core
In a computer simulation, the position of the $2\text{s}$ electron of a hydrogen atom is measured $8000$ times. Of the measurements, $421$ land inside the nodal sphere of the $2\text{s}$ orbital. Based on the measurements, estimate the probability with which the $2\text{s}$ electron is detected inside the nodal sphere. Give the answer as a percentage.
Answer
$5.26\,\%$
Show the solution
Given
measurements $N = 8000$, hits inside the nodal sphere $k = 421$
Asked
the probability $P$ of detecting the $2\text{s}$ electron inside the nodal sphere, as a percentage
Principle
According to the probability interpretation, only probabilities can be calculated from the wave, and from measurements made in many atoms in the same state the probability is estimated as the share of the hits that land in the region. The more measurements there are, the more accurate the estimate.
Formula
the estimate of the probability is $P = \dfrac{k}{N}$
Conversions
The share is converted to a percentage by multiplying by one hundred: $1 = 100\,\%$.
Substitution
$P$$= \dfrac{421}{8000}$$= 0.052\,625$$= 5.2625\,\%$Answer
$5.26\,\%$
Three significant figures like the number of hits $421$.
Reasonableness
About every twentieth measurement lands inside the nodal sphere, since $8000 : 20 = 400$, which is close to the number $421$. The share is small, because the radius of the nodal sphere is only about $0.11\,\text{nm}$ and most of the probability of the $2\text{s}$ electron is outside it.
- Exercise 5-3
The position of the $1\text{s}$ electron of a hydrogen atom is measured $3000$ times, and the $90\,\%$ boundary of the $1\text{s}$ orbital is drawn in the picture.
How many measurements do you expect to land outside the boundary?
Answer
$300$
Show the solution
Given
measurements $N = 3000$, boundary of the orbital $90\,\%$
Asked
the expected number $k_\text{out}$ of measurements that land outside the boundary
Principle
The boundary is drawn so that the electron is detected inside it with probability $0.90$, so the probability of landing outside is $1 - 0.90 = 0.10$. In many measurements the share of those landing outside is approximately this probability, $P \approx \tfrac{k}{N}$.
Formula
$k_\text{out} = (1 - p)\,N$
Conversions
$p = 90\,\% = 0.90$
Substitution
$k_\text{out}$$= (1 - 0.90) \cdot 3000$$= 0.10 \cdot 3000$$= 300$Answer
about $300$ measurements; the number is an expected value, and the real number varies from one series to another.
Reasonableness
Every tenth measurement lands outside the boundary, and a tenth of $3000$ is $300$. If the boundary were a wall, there would be no hits outside, but the boundary of an orbital is only an agreed probability boundary.
- Exercise 5-4
The electron of a hydrogen atom is in the $3\text{p}$ state.
At least how much energy is needed to release the electron from the atom? Give the answer in electronvolts.
Answer
$1.51\,\text{eV}$
Show the solution
Given
electron in the $3\text{p}$ state, so $n = 3$; energy levels of the hydrogen atom $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$
Asked
the smallest energy $E_\text{i}$ needed for release, in electronvolts
Principle
The electron is free of the atom when its energy is at least $0$. In the hydrogen atom the energy of a state depends only on the principal quantum number, so the energy of the $3\text{p}$ state is the energy of the shell $n = 3$, and the energy needed is the difference $0 - E_3$.
Formula
$E_\text{i}$$= 0 - E_3$$= \dfrac{13.6\,\text{eV}}{3^2}$Conversions
None are needed, because the energy levels and the asked energy are in electronvolts.
Substitution
$E_\text{i}$$= \dfrac{13.6\,\text{eV}}{9}$$= 1.5111\,\text{eV}$Answer
$1.51\,\text{eV}$
Three significant figures like the energy $13.6\,\text{eV}$.
Reasonableness
On the shell $n = 3$ the electron is farther from the nucleus and more weakly bound than in the ground state, so release takes only a ninth of the ground-state energy $13.6\,\text{eV}$. The energy is positive, because it is brought into the atom.
- Exercise 5-5
The electron of a hydrogen atom is in the shell $n = 5$. What is its energy in electronvolts?
Answer
$-0.544\,\text{eV}$
Show the solution
Given
shell $n = 5$; energy levels of the hydrogen atom $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$
Asked
the energy $E_5$ of the electron in electronvolts
Principle
In the hydrogen atom the energy of the state of the electron depends only on the principal quantum number, and the energy levels are given by the formula. The energy is negative, because a bound electron needs energy to be released.
Formula
$E_5 = -\dfrac{13.6\,\text{eV}}{5^2}$
Conversions
None are needed, because the energy is asked in electronvolts.
Substitution
$E_5$$= -\dfrac{13.6\,\text{eV}}{25}$$= -0.544\,\text{eV}$Answer
$-0.544\,\text{eV}$
Three significant figures like the energy $13.6\,\text{eV}$.
Reasonableness
The energy is close to zero, because the electron of the shell $n = 5$ is weakly bound. It is greater than $E_4 = -0.85\,\text{eV}$, as it should be, because the energy increases as the number of the shell increases.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 5-6Core
The electron of a hydrogen atom moves from the $3\text{p}$ orbital to the $1\text{s}$ orbital, and the atom emits a photon.
What is the wavelength of the photon? Give the answer in nanometres.
Answer
$103\,\text{nm}$
Show the solution
Given
initial state $3\text{p}$, so $n = 3$; final state $1\text{s}$, so $n = 1$; $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$, $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$, $c = 2.998 \cdot 10^{8}\,\text{m/s}$
Asked
the wavelength $\lambda$ of the photon in nanometres
Principle
Energy is conserved: the energy of the atom decreases by the difference between the energies of the states, and the same energy leaves as a photon, whose energy is $hf = \tfrac{hc}{\lambda}$. In the hydrogen atom the energy of a state depends only on the principal quantum number.
Formula
$\dfrac{hc}{\lambda} = E_3 - E_1$, so $\lambda = \dfrac{hc}{E_3 - E_1}$
Conversions
The energy difference in electronvolts is
$E_3 - E_1$$= -\tfrac{13.6\,\text{eV}}{9} + 13.6\,\text{eV}$$= 12.089\,\text{eV}$Substituting $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$:
$E_3 - E_1$$= 12.089 \cdot 1.602 \cdot 10^{-19}\,\text{J}$$= 1.9366 \cdot 10^{-18}\,\text{J}$Finally $1\,\text{nm} = 10^{-9}\,\text{m}$.
Substitution
First
$hc$$= 6.626 \cdot 10^{-34}\,\text{J}\,\text{s} \cdot 2.998 \cdot 10^{8}\,\text{m/s}$$= 1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}$and then
$\lambda$$= \dfrac{1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}}{1.9366 \cdot 10^{-18}\,\text{J}}$$= 1.0257 \cdot 10^{-7}\,\text{m}$$= 102.57\,\text{nm}$Answer
$103\,\text{nm}$
Three significant figures like the energy $13.6\,\text{eV}$.
Reasonableness
The energy of the photon is about $12\,\text{eV}$, more than the energy $10.2\,\text{eV}$ of the transition $2\text{p} \to 1\text{s}$, so the wavelength must be shorter than the $122\,\text{nm}$ of that transition. The radiation is ultraviolet, like all the lines of hydrogen that end in the ground state.
- Exercise 5-7Core
An argon atom has $18$ electrons, and in the ground state they fill the orbitals in the order $1\text{s}$, $2\text{s}$, $2\text{p}$, $3\text{s}$, $3\text{p}$.
How many electrons are in the outermost shell of argon?
Answer
$8$
Show the solution
Given
argon has $18$ electrons; filling order $1\text{s}$, $2\text{s}$, $2\text{p}$, $3\text{s}$, $3\text{p}$; at most two electrons per orbital
Asked
the number of electrons $N_\text{outer}$ in the outermost shell
Principle
In the ground state the electrons fill the orbitals starting from the lowest energy. At most $2n^2$ electrons fit in the shell $n$, so first the shells $n = 1$ and $n = 2$ fill, and the remaining electrons are in the shell $n = 3$, which is the outermost shell of argon.
Formula
$N_\text{outer} = N - 2 \cdot 1^2 - 2 \cdot 2^2$
Conversions
None are needed, because a count is asked.
Substitution
$N_\text{outer} = 18 - 2 - 8 = 8$
Answer
$8$
Electrons; the result is an exact integer.
Reasonableness
On the shell $n = 3$ the electrons are in the $3\text{s}$ and $3\text{p}$ subshells: $2 + 6 = 8$, so both subshells are full. The configuration $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6$ contains $18$ electrons. Full outermost s and p subshells fit the fact that argon is a noble gas that hardly forms bonds.
- Exercise 5-8
Two boundaries are drawn for the $1\text{s}$ orbital of a hydrogen atom: the $50\,\%$ boundary, whose radius is $0.071\,\text{nm}$, and the $90\,\%$ boundary, whose radius is $0.14\,\text{nm}$. The position of the electron is measured $2400$ times.
How many measurements do you expect to land between the boundaries?
Answer
$960$
Show the solution
Given
measurements $N = 2400$; $50\,\%$ boundary $0.071\,\text{nm}$ and $90\,\%$ boundary $0.14\,\text{nm}$
Asked
the expected number $k$ of measurements that land between the boundaries
Principle
The $50\,\%$ boundary lies entirely inside the $90\,\%$ boundary, so the probability $0.90 - 0.50 = 0.40$ remains in the shell between the boundaries. In many measurements the share of those landing in a region is approximately the probability of the region. The radii of the boundaries are not needed.
Formula
$k = (p_{90} - p_{50})\,N$
Conversions
$p_{90} = 90\,\% = 0.90$ and $p_{50} = 50\,\% = 0.50$
Substitution
$k$$= (0.90 - 0.50) \cdot 2400$$= 0.40 \cdot 2400$$= 960$Answer
about $960$ measurements; the number is an expected value, and the real number varies from one series to another.
Reasonableness
Half, that is $1200$ measurements, land inside the $50\,\%$ boundary and $2160$ measurements inside the $90\,\%$ boundary; their difference is $960$. There are fewer hits between the boundaries than inside the inner boundary, although the shell has a greater volume, because the cloud thins out outwards.
- Exercise 5-9
The electron of a hydrogen atom is in the shell $n = 2$.
What is the greatest wavelength at which a photon can still release the electron from the atom? Give the answer in nanometres.
Answer
$365\,\text{nm}$
Show the solution
Given
electron in the shell $n = 2$; $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$, $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$, $c = 2.998 \cdot 10^{8}\,\text{m/s}$
Asked
the greatest wavelength $\lambda_\text{max}$ at which a photon releases the electron, in nanometres
Principle
The energy of the photon must be at least the energy $0 - E_2$ needed for release. The energy of a photon, $\tfrac{hc}{\lambda}$, decreases as the wavelength increases, so the greatest suitable wavelength corresponds to exactly this energy.
Formula
$\dfrac{hc}{\lambda_\text{max}}$$= 0 - E_2$so $\lambda_\text{max} = \dfrac{hc}{-E_2}$
Conversions
$-E_2$$= \tfrac{13.6\,\text{eV}}{4}$$= 3.40\,\text{eV}$Substituting $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$:
$-E_2$$= 3.40 \cdot 1.602 \cdot 10^{-19}\,\text{J}$$= 5.4468 \cdot 10^{-19}\,\text{J}$Finally $1\,\text{nm} = 10^{-9}\,\text{m}$.
Substitution
First
$hc$$= 6.626 \cdot 10^{-34}\,\text{J}\,\text{s} \cdot 2.998 \cdot 10^{8}\,\text{m/s}$$= 1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}$and then
$\lambda_\text{max}$$= \dfrac{1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}}{5.4468 \cdot 10^{-19}\,\text{J}}$$= 3.6470 \cdot 10^{-7}\,\text{m}$$= 364.70\,\text{nm}$Answer
$365\,\text{nm}$
Three significant figures like the energy $13.6\,\text{eV}$.
Reasonableness
The energy $3.4\,\text{eV}$ is slightly above the energy of a photon of violet light, so the wavelength is just below the visible range, near ultraviolet. Visible light with a longer wavelength does not release an electron from the shell $n = 2$.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 5-10
An ultraviolet photon with a wavelength of $97.3\,\text{nm}$ hits a hydrogen atom in the ground state, and the atom absorbs it.
How many orbitals does the shell to which the electron moves have?
Answer
$16$
Show the solution
Given
wavelength of the photon $\lambda = 97.3\,\text{nm}$; atom in the ground state, $n = 1$; $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$, $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$, $c = 2.998 \cdot 10^{8}\,\text{m/s}$
Asked
the number of orbitals in the shell $n$ to which the electron moves
Principle
An atom can absorb a photon only if the energy of the photon is exactly the difference between the energies of two states. The energy of the photon gives the energy of the upper state, $E_n = E_1 + \tfrac{hc}{\lambda}$, from that the number of the shell $n$, and the shell $n$ has $n^2$ orbitals.
Formula
$E_n = E_1 + \dfrac{hc}{\lambda}$, so
$n$$= \sqrt{\dfrac{13.6\,\text{eV}}{-E_n}}$and the number of orbitals is $n^2$
Conversions
$\lambda$$= 97.3\,\text{nm}$$= 97.3 \cdot 10^{-9}\,\text{m}$With
$hc$$= 6.626 \cdot 10^{-34}\,\text{J}\,\text{s} \cdot 2.998 \cdot 10^{8}\,\text{m/s}$$= 1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}$the energy of the photon is
$\tfrac{hc}{\lambda}$$= \tfrac{1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}}{97.3 \cdot 10^{-9}\,\text{m}}$$= 2.0416 \cdot 10^{-18}\,\text{J}$It is converted to electronvolts by substituting $1\,\text{J} = \tfrac{1}{1.602 \cdot 10^{-19}}\,\text{eV}$:
$\tfrac{2.0416 \cdot 10^{-18}}{1.602 \cdot 10^{-19}}\,\text{eV}$$= 12.744\,\text{eV}$Substitution
$E_n$$= -13.6\,\text{eV} + 12.744\,\text{eV}$$= -0.856\,\text{eV}$$n$$= \sqrt{\dfrac{13.6\,\text{eV}}{0.856\,\text{eV}}}$$= \sqrt{15.9}$$= 3.99$$\approx 4$so there are $4^2 = 16$ orbitals
Answer
$16$
Orbitals (shell $n = 4$: $4\text{s}$, three $4\text{p}$, five $4\text{d}$ and seven $4\text{f}$); $n$ is an integer, so the rounding $3.99 \to 4$ is reliable.
Reasonableness
The energy of the shell $n = 4$, $-\tfrac{13.6\,\text{eV}}{16} = -0.85\,\text{eV}$, equals the calculated $-0.856\,\text{eV}$ to the precision of three figures of the wavelength. The energies of the shells $n = 3$ ($-1.51\,\text{eV}$) and $n = 5$ ($-0.54\,\text{eV}$) are clearly farther away, so no other shell fits.
- Exercise 5-11
In light atoms, the electrons fill the orbitals in the ground state in the order $1\text{s}$, $2\text{s}$, $2\text{p}$, $3\text{s}$, $3\text{p}$, and at most two electrons fit in each orbital. In an atom of a certain element the $3\text{p}$ subshell is exactly half full.
How many electrons does the atom have?
Answer
$15$
Show the solution
Given
filling order $1\text{s}$, $2\text{s}$, $2\text{p}$, $3\text{s}$, $3\text{p}$; at most two electrons per orbital; the $3\text{p}$ subshell is half full
Asked
the number of electrons $N$ in the atom
Principle
In the ground state the electrons fill the orbitals starting from the lowest energy, so electrons go to the $3\text{p}$ subshell only when $1\text{s}$, $2\text{s}$, $2\text{p}$ and $3\text{s}$ are full. An s subshell has one orbital and a p subshell three, so they hold $2$ and $6$ electrons. A half-full $3\text{p}$ subshell contains half of its maximum number.
Formula
$N$$= N_\text{1s} + N_\text{2s} + N_\text{2p} + N_\text{3s} + \dfrac{N_\text{3p,max}}{2}$Conversions
None are needed, because a count is asked.
Substitution
$N$$= 2 + 2 + 6 + 2 + \dfrac{6}{2}$$= 2 + 2 + 6 + 2 + 3$$= 15$Answer
$15$
Electrons (the element is phosphorus, configuration $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^3$); the result is an exact integer.
Reasonableness
The full shells $n = 1$ and $n = 2$ hold $2 + 8 = 10$ electrons, and the shell $n = 3$ has $2 + 3 = 5$ electrons, $15$ altogether. In the periodic table the atomic number $15$ is phosphorus, which is in the middle of the third period, as a half-full $3\text{p}$ subshell leads one to expect.
- Exercise 5-12
The electron of a hydrogen atom is in the $2\text{s}$ state, and its position is measured $5000$ times. The $90\,\%$ boundary of the $2\text{s}$ orbital is drawn in the picture. On average $5.3\,\%$ of the measurements land inside the nodal sphere, and this inner part belongs practically entirely to the region of the orbital.
How many measurements do you expect to land in the region of the orbital outside the nodal sphere?
Answer
$4240$
Show the solution
Given
measurements $N = 5000$; boundary of the orbital $90\,\%$; $5.3\,\%$ of the measurements land inside the nodal sphere, and this inner part belongs practically entirely to the region of the orbital
Asked
the expected number $k$ of measurements that land in the region of the orbital outside the nodal sphere
Principle
The probability $0.90$ of the region of the orbital divides into two parts: the inner part of the nodal sphere and the shell outside the nodal sphere. Since the inner part belongs entirely to the region, the probability of the outer part is the difference $0.90 - 0.053$. In many measurements the share of those landing in a region is approximately the probability of the region.
Formula
$k = (p - p_\text{in})\,N$
Conversions
$p = 90\,\% = 0.90$ and $p_\text{in} = 5.3\,\% = 0.053$
Substitution
$k$$= (0.90 - 0.053) \cdot 5000$$= 0.847 \cdot 5000$$= 4235$Answer
about $4240$ measurements; the number is an expected value that a real series fulfils only approximately, and the share of the inner part, $5.3\,\%$, limits the precision to about two or three significant figures.
Reasonableness
Altogether about $0.90 \cdot 5000 = 4500$ measurements land in the region, and of these about $0.053 \cdot 5000 = 265$ inside the nodal sphere; the difference $4235$ is clearly the largest part. This fits the fact that the most common distance of the $2\text{s}$ electron is outside the nodal sphere.
Photographs
- ESR-STM at QNS in Ewha - front view. Rickinasia. source, CC BY-SA 4.0
