Mod 8 U2A

Question 1

What was the method of JJ Thomson’s experiment

Answer 1

1. Acceleration of electrons from cathode (negative) to the anode (positive)
2. Deflection of electrons by magnetic fields. Cathode ray was deflected with a magnetic field allowed radius of the curvature of the path to be determined. qvB = mv²/r -> q/B = v/Br
3. Deflection of electrons by both magnetic and electric fields. Electric field was applied to bring cathode ray into a straight path. Effectively balancing electric and magnetic forces. v = E/B
4. Using size of deflection and measured values of E and B q/m could be determined

Question 2

What were the results of Thomson’s experiment

Answer 2

• Thomson measured q/m of around 1.5 x 10¹¹ C/kg
• Much larger than charge to mass ratio of all other known charged particles (ions) hence indicating discovery of a new particle
• cloest value was that of hydrogen atoms which was 2000 times smaller
• Thomson found q/m for cathode rays was the same for different gases and different electrode materials providing evidence that electrons are present in all matter.

Question 3

Conclusion of Thomson’s experiment

Answer 3

• The electron is a particle which is: fundamental, subatomic and negatively charged
• Cathode rays are streams of electrons flowing from the cathode to the anode
• Thomson was only able to measure q/m, could not seperate q or m. This was accomplished later by Millikan.

Question 4

Outline the production of cathode rays in cold discharge tubes

Answer 4

1. A small trace of ions and free electrons are present in the residual gas
2. A high voltage is applied between the cathode (negative) and the anode (positive), creating a strong electric field in the tube
3. The strong electric field accelerates the ions towards the cathode and electrons towards the anode (cathode rays)
4. As the ions and electrons accelerate through the tube they collide with other molecules and ionise them, releasing more free electrons
5. As the electrons accelerate through the discharge tube they collide with the air molecules as a result light is produced

Question 5

What are the three consequences of an electron (cathode rays) colliding with an an atom or molecule

Answer 5

1. Low energy collision. Electron is deflected (electrical resistance)
2. High energy collision. Some of the kinetic energy the electrons acquired as a result of the high potential difference dk = qV is converted to light energy (making cathode rays visible)
3. Very high energy collision. The electron ionises the atom or molecule, resulting in an ion and extra free electron.

Question 6

Original cathode ray experiments and their conclusions

Answer 6

1. Maltese cross, travel in straight lines
2. Paddle wheel, have momentum and hence mass (inaccurate and was actually due to the radiometric effect)
3. Deflection by magnetic fields, must be charged
4. Cathode rays carry negative charge, negatively charged

Question 7

Diagram of deflection of magnetic fields in Thomson’s experiment

Question 8

Method of Millikan’s oil drop experiment

Answer 8

1. Drops of oil are sprayed into a region above a pair of parallel plates
2. The drops become ionised either through friction with the nozzle of the spray or by illuminating them with X-rays
3. Drops can fall into the region between the plates. This region contains an electric field set-up by a voltage across the plates
4. Once between the plates the drops can be viewed through a microscope. The drops can be accelerated up or down by adjusting teh voltage on the plates, and hence the electric field
5. The voltage is adjusted such that the drops are suspended (qE = mg, q = mg/E)

Question 9

Results of Millikan oil drop experiment

Answer 9

1. q = mg/E (after balancin qE and mg)
2. E = V/d (known) and g = 9.8ms^-2 (just need mass)
3. Mass of oil drop was calculated using the oil drops terminal velocity, when F_net = 0 since air resistance balances out gravitational force, and measured their diamter to find mass
4. Then use mass of oil drop to calculate the charges of many oil drops and worked out the charges of the oil drops were always an integer multiple of some constant
5. Constant was charge of an electron, this value was then used with Thomson’s q/m to work out the mass of an electron.

Question 10

Outline Thomson’s Plum Pudding model of the atom (Disproved by Gieger Marsden)

Answer 10

• Thomson suggested that an atom is a sphere of positive charge in which electrons are embedded like raisins in a plum pudding
• No protons in the PP model as they had not yet been discovered

Question 11

Outline the method of the Geiger-Marsden experiment

Answer 11

1. A narror beam of alpha particles from a radioactive source was aimed at a thin foil of metal such as gold
2. A zinc sulfide screen was used to detect the position of the deflected alpha particles (Zinc sulfide fluoresces when struck by alpa particles)

Question 12

Theoretical prediction of the Geiger-Marsden experiment

Answer 12

• On the basis of the Thomson model of the atom Rutherford expected the alpha particles to pass through the foil virtually undeflected
• Based on Thomson’s model, Rutherford expected that the gold atoms would not affect the positive alpa particles because in gold atoms there is no concentation of of charge and mass large enough to deflect the alpha particles

Question 13

Why doesn’t Rutherford’s model work

Answer 13

• Cannot explain why atoms are stable
• Cannot explain the aborption and emission spectra of atoms

Question 14

Experimental results of the Geiger Marsden experiment

Answer 14

• Most of the alpha particles passed through undeflected as expected
• Some were very significantly deflected
• A few bounced back straight towards the source

Question 15

Conclusion of the Geiger Marsden experiment

Answer 15

• Rutherford concluded that there must be a small, massive, positively charged core in the atoms that causes the alpha particles to deflect
• Analysis was:
• No deflection, most particles, alpha particles do not interact with the nucleus
• Significant deflection, some particles, alpha particles interact weakly with the nucleus
• Reflected to source, very few particles, alpha particles pass close to nucleus and interact strongly

Question 16

Outline details of Rutherfords model (does not work)

Answer 16

• The nucleus is a central region that is very small relative to the total size of an atom
• The nucleus contains virtually all of the mass and all the positive charge of the atom
• The positive charge of the nucleus is to be balanced by the negative charge on the electrons outside the nucleus to maintain electrical neutrality
• The electrons orbit around the nucleus in uniform circular motion.

Question 17

Explain why Rutherfords model cannot explain why atoms are stable

Answer 17

• Electrons orbit nucleus in uniform circular motion which requires centripetal force
• Hence the electrons are accelerating
• According to Maxwell’s theory accelerating charges emit EM waves
• Meaning electrons would lose energy by EMR
• Losing energy would cause them to orbit increasingly close to the nucleus, with increasing frequency, ultimately leading to collapse of the atom
• This does not happen as atoms are stable

Question 18

Explain why Rutherford’s model cannot explain the absorption and emission spectra of atom

Answer 18

• At the time of Rutherfords work scientists had characterised the emission and absorption spectra of different element
• Hydrogen and other elements absorb only specifc wavelengths and emit light only at the wavelengths they absorb.
• Rutherford’s model had no structure for the electrons, they were free to orbit at any radius, hence they should be able to absorb and emit radiation at any wavelength

Question 19

What is the main cause for Rutherford’s model being unable to explain the stability of atoms and absorption and emission spectra

Answer 19

• Problems arise due to the classical nature of Rutherford’s model.
• It treats the motion of the electrons using Newton’s laws and makes predictions based on Maxwell’s theories
• These are both classical physics theories which do not work at a subatomic level, this requires quantum physics.

Question 20

What are Bohr’s postulates

Answer 20

1. Only certain electron orbits are stable
2. Radiation is absorbed or emitted when electrons transition from one stable orbit to another (higher energy they absorb energy, if they move to a lower energy they emit energy), difference in energy is E_i - E_f = hf
3. The stable electrons have quantised angular momentum L_n = mvr = nh/2pi

Question 21

Outline Rutherford’s identification of the proton

Answer 21

1. Rutherford observed that hydrogen nuclei were produced when alpha particles were fired at nitrogen
2. From this he concluded that nitrogen nuclei (and hence all nuclei) contained hydrogen within them
3. The nuclear reaction that was occuring in his experiment was: ¹⁴₇N + alpha -> ¹⁷₈O + ¹₁H
4. Rutherford proposed that the nucleus of hydrogen consisted of a fundamental building block which made up all other nuclei
5. He named this particle the proton in 1920

Question 22

Outline method of James Chadwick’s disovery of Neutrons

Answer 22

• Polonium is used as an alpha source and the emitted alphaa particles are directed at a piece of beryllium
• When struck the berryllium emitted mysterious “neutral rays” which were highly penetrating
• Paraffin wax (which contains a lot of hydrogen) was placed in the path of the mysterious radiation. When the rays hit the paraffin, they knocked protons out of it
• Due to their charge, the protons were able to be detected by a Geiger counter and their energy and speed could be characterised

Question 23

Explain the conclusion of Chadwick’s experiment

Answer 23

• The following reaction was taking place between the alpha particles and the berrylium which resulted in the release of neutrons: ⁹₄Be + alpha -> ¹²₆C + ¹₀n
• Neutrons travelled at high speed and collided with the nuclei in the target
• Collisions caused the nuclei to be knocked out
• Nuclei was easily detected as they were charged their mass was known and the speed could be measured
• This allowed energy and speed to be measured
• Using conservation of energy and momentum, Chadwick showed that whatever was colliding with the nuclei must have been an uncharged particle with a mass close to that of the proton i.e. a neutron

Question 24

What is the emission spectrum of a chemical element

Answer 24

• This is the spectrum of frequencies of electromagnetic radiation emitted due to an electron undergoing a transition from a high energy state to a lower energy state
• E_i - E_f = hc/lamda

Question 25

What is the absorption spectrum of chemical elements

Answer 25

• This is the spectrum of frequencies of electromagnetic radiation absorbed due to an electron undergoing a transition from a low energy state to a higher energy state
• E_i - E_f = hc/lamda

Question 26

Which problems of Rutherford’s model did Bohr’s model fix

Answer 26

• Bohr’s atomic model was stable by definition
• Provided an explanation of the atomic spectrum of hydrogen

Question 27

Outline how Bohr was able to explain the existence of line spectra

Answer 27

1. Electrons exist in stable orbits only which have quantised angular momentum and energy, E_n = -13.6eV/n², L_n = nh/2pi
2. Electrons will absorb and emit photons when they transition from one stable orbit to another
3. Since the energy of the stable orbit is fixed, the energy difference and hence photon energies will be fixed
4. Wavelengths absorbed will be fixed

Question 28

General flowchart for extended response question

Answer 28

1. Impact on scientific thinking (what is it about) Describe
2. Change (before and after) Evaluate
3. How did ….. improve our understanding of ….. (Evaluate)

Question 29

Which quantum number does the Balmer series correspond to

Answer 29

2

Question 30

What is significance of the Balmer series, any number n = 3-6 to 2

Answer 30

This transition of any n value to n = 2 produces wavelengths in the visible light spectrum

Question 31

Formular for Balmer series

Answer 31

1/lambda = R_H (1/2² - 1/n²)

Question 32

Derive Rydberg equation from Bohr’s model

Answer 32

1. E_n = -13.6eV/n²
2. If an electron moves from one stable orbit n_i to a lower stable orbit n_f it must emit the energy difference as a photon, the photon energy will be E = -13.6eV (1/n_i²- 1/n_f²)
3. 1/lamda = E/hc
4. combine two expressions 1/lamda = 13.6eV/hc (1/n_f² - 1/n_i²)
5. 13.6eV/hc = 1.097 x 10⁷m^-1 = Rydbergs constant
6. Thus Bohr’s model allows the calculation of the energy of each stable orbit of hydrogen

Question 33

What were the main limitations of Bohr’s model

Answer 33

• Why the stable orbits were stable (it said they were but didn’t explain why)
• The spectra of atoms with more than one electron
• The relative intensity of spectral lines
• The Zeeman effect
• Hyperfine splitting

Question 34

What were the underlying reasons for Bohr’s model being unable to explain certain things

Answer 34

• It uses one quantum number n to describe the stable orbits. The correct quantum model requires four quantum numbers
• Since it was a simple model, it could only explain the simplest atom: Hydrogen

Question 35

Explain why Bohr’s analysis did not work for atoms that have more than one electron

Answer 35

It did not take into account interactions between electrons

Question 36

What is the relative intensity of spectral lines

Answer 36

• There are different types of spectral lines
• ’s’ sharp lines
• ‘p’ primary lines
• ‘d’ diffuse lines
• ‘f’ fine lines
• Bohr’s model could not account for these

Question 37

What is the Zeeman effect

Answer 37

• When spectral lines were recorded with a discharge tube placed in a magnetic field, the recorded spectral lines were split into several finely seperated lines
• Three spectral lines (instead of just one) indicate three different transitions for the electron
• There must be three stable orbits for the electron to go to instead of just one as predicted by Bohr
• The energy of these orbits must be affected by the magnetic field

Question 38

Explain Hyperfine spectral lines in relation to Bohr’s model

Answer 38

• When spectroscopic techniques were improved and used to examine spectral lines of hydrogen, it was revealed that many of the lines in the Balmer series and others were not single lines at all
• Instead, each line was actually a pair of lines spaced very close together. This is known as hyperfine splitting
• This indicated the presence of multiple stable orbits where Bohr’s model predicted one.
• Closely spaced lines indicate two closely spaced energy states for the electron

Question 39

What is de Broglie’s hypothesis

Answer 39

• Postulated that because photons have a dual wave/particle nature, all forms of matter have wave as well as particle properties
• Electrons have a dual wave/particle nature
• In analogy with photons having a wavelength, he proposed that the wavelength of a particle is related to momentum by:
• lamda_particle = h/p = h/mv (combines waves and particle properties)
• Accompanying every electron is a wave (not an EM wave), which guides or pilots the electrons through space. We can then picture matter particles in the same way as we picture photon

Question 40

What was the main experimental evidence of matter waves

Answer 40

The Davisson-Germer experiment

Question 41

Outline procedure of Davisson and Germer experiment

Answer 41

• Electrons were emitted by a heated filament (cathode) and accelerated towards an anode by a voltage (54V in this case).
• The electrons passed through a hole in the anode and were directed at a piece of metal
• A detector was used to measure the number of electrons deflected at different angles
• The detector was rotated around the metal to measure the number of electrons as a function of angle

Question 42

Outline how results of Davisson Germer experiment were used to calculate the wavelength of the electron

Answer 42

• Scattered electrons exhibited maxima and minima which shows constructive and destructive interference (wave property)
• They identified the peak at 50^o corresponded to diffraction and constructive interference
• Hence they could use the diffraction equation to calculate wavelength:
• 2dsin(theta) = m(lamda), where:
• d is the atomic spacing
• Theta = 90^o - phi/2
• M is the diffracted order (m = 1 as it is the first constructive interference peak away from 0^o)

Question 43

How is de Broglie wavelength of an electron whose KE is x Volts determined

Answer 43

• Using KE = mv² find velocity
• Using lamda = h/mv find wavelength

Question 44

Why are standing waves called ‘standing’ waves

Answer 44

Because the nodes (troughs) and antinode (peaks) of the wave appear stationary

Question 45

How did de Broglie resolve the problem with the stability of electron orbits in the Bohr atom

Answer 45

• de Broglie used interference to resolve the problem, postulating that stable orbits are standing waves
• de Broglie proposed that:
• Electrons can orbit an atomic nucleus only if the circumference of the orbit is a whole number of electron wavelengths, so that it forms a standing wave
• n(lamda) =2pi(r)

Question 46

Explain n(lamda) = 2pi(r)

Answer 46

• n is an integer, indicating the number of wavelengths that fit around the circumference
• 2pi(r) is the circumference
• If the electron wave travels around the circular orbit it will return to the same place and superimpose with itself, leading to interference
• if it travels a whole number of wavelengths (n) around the circumference, it will be in phase with itself at every position, and will form a standing wave
• if it doesn’t travel a whole number of wavelengths around the circumference it will eventually be out of phase with itself and will not form a standing wave

Question 47

Derive mvr = nh/2pi

Answer 47

• n(lamda) = 2pi(r), condition of the stability of an electron orbit
• lamda = h/mv, de Broglie’s expression
• n(h/mv) = 2pi(r)
• mvr = nh/2pi

Question 48

What probelms did the Bohr-de Broglie model still face

Answer 48

• It was essentially one dimensional
• It did not explain how the electrons would be distributed amongst different standing waves
• It could not explain many features of atomic spectra (spectra of larger atoms, Zeeman effect, hyperfine splitting, relative intensity of spectral lines)

Question 49

What does the Schrodinger equation describe

Answer 49

How particles behave given their wave nature

Question 50

What are the three quantum numbers and what do they do

Answer 50

• n, l and m_l
• The quantum number n (shell) is called the principal quantum number. The values of n can range from 1 to infiniti, this is the quantum number Bohr introduced
• The quantum number l (subshell) is called the quantum orbital number. The values of l for a given n can range from 0 to n - 1
• The quantum number m_l is called the orbital magnetic quantum number. The values of m_l for a given l can range from -l to +l

Question 51

How many quantum numbers are required to describe the quantum states of an electron in an atom

Answer 51

• Four: n, l, m_l, m_s
• The first three are the quantum numbers that describe the orbitals obtained from solving Schrodinger’s equation
• The fourth quantum number describes spin
• Every orbital corresponds to two quantum states one for each spin

Question 52

What are the two states for the spin magnetic quantum number

Answer 52

• The spin up state is labelled m_s = 1/2
• The spin down state is labelled m_s = -1/2

Question 53

What is the Pauli Exclusion Principle

Answer 53

• No two electrons can ever be in the same quantum state: that is no two electrons in the same atom can have the same set of quantum numbers, n, l, m_l, m_s
• Hence the first electron goes to the lowest energy state
• The next electron goes to the next lowest energy state
• This continues until all the electrons are assigned a quantum state

Question 54

What is the Heisenberg uncertainty principle

Answer 54

• Idea that it is physically impossible to simultaneously measure the exact position and exact momentum of a particle
• Photon thing examing an electron, photon must bounce off electron and go up the microscope

Question 55

What is mass defect and why does it occur

Answer 55

• The difference between the sum of all individual nucleons (protons, neutrons and electrons) and the mass of the atom
• Because mass of the nucleons is partially converted to binding energy to hold the nucleus together

Question 56

Define binding energy

Answer 56

• The binding energy of a nucleus is defined as the total work required to overcome the strong force and seperate its nucleons

Question 57

What effect does doing work equal to the binding energy have on the nucleons

Answer 57

• Doing work will increase their total energy from the negative value they have as bound nucelons to zero for a free stationary nucleon
• By increasing their total energy, the mass of the nucleons will increase from its mass inside the nucleus to their rest mass m₀
• Hence by seperating the nucleons their mass will increase by an amount equal to the mass defect
• Binding energy corresponds to mass defect throug dmc²

Question 58

Steps to calculate the binding energy of a nucleus

Answer 58

1. Calculate the mass of the constituents
2. Calculate the mass defect by subtracting the mass of the nucleus or atom from the mass of the consitutents
3. Calculate the binding energy using E = mc²
4. If using nucleus only do not include electrons

Question 59

What does binding energy per nucleon reveal

Answer 59

• The stability of an atom

Question 60

Outline Hubble’s discovery of universe expansion

Answer 60

1. Leavitt found relationship between the luminosity of Cephied variables and their period of oscillation from bright to dim
2. Using relationship distances in space can be measured using inverse sqaure law
3. Hubble observed Cepheid variables in the andromeda and used their period