PARTICLE PHYSICS AND QUANTUM PHYSICS A Flashcards
1
Q
Describe the process of pair production.
A
-when a gamma ray photon interacts with a nucleus
-producing a particle anti-particle pair
- energy of photon must be > combined rest energies of particles
- nucleus gains backwards momentum to conserve momentum
-extra energy is split evenly into the kinetic energies of the particles
2
Q
Describe the process of ANNIHILATION.
A
- when a particle and it’s corresponding antiparticle meet and destroy each other
-mass is converted into energy and split evenly in the form of two gamma photons travelling in opposite directions to conserve momentum - 2RE + KE = 2E
3
Q
Draw the diagram for beta-minus decay
A
n –(W- boson)–> p + e- + electron anti-neutrino
4
Q
Draw the diagram for beta-plus decay
A
p –(W+ boson)–> n + e+ + electron neutrino
5
Q
Draw the Feynman Diagram for electron capture.
A
p + e- –(W+ boson)–> n + electron neutrino
6
Q
Give the Feynman diagram for electron-proton collision.
A
p + e- –(W- boson)–> n + electron neutrino
7
Q
Sketch the Feynman Diagram of EMF repulsion.
A
two protons on either side, with a virtual photon connecting them.
Must be labeled with gamma symbol, and have no arrows on the virtual photon as it is chargeless
8
Q
Give the equation for specific charge. Now, without working, explain how you can tell which has the greater magnitude of specific charge: an electron or a proton.
A
specific charge = Q/m. Since specific charge is directly proportional to 1/m, an electron will certainly have the larger magnitude of specific charge.
9
Q
A sample of bromine gas contains a mixture of 2 isotopes. Each molecule has 2 bromine atoms. The experiment finds that the bromine molecules contain 158, 160 and 162 nucleons. Bromine has an atomic number of 35. Find the number of neutrons for the isotope with the greater nucleon number.
A
Let x = nucleon number of isotope 1 and y = nucleon number of isotope 2. y > x.
Form 2 eqns: x + y =160, 2x = 158.
x = 79
y = 81
81-35 = 46.
10
Q
State the magnitude of charge of an electron. Don’t forget significant figures and include units.
A
1.60 x 10^-19 C
11
Q
State the mass of a nucleon. to 3 sig fig.
A
1.67 x 10 ^ -27 kg
12
Q
State the value of planck’s constant and give units.
A
6.63 x 10 ^-34 Js
13
Q
Describe in full detail, the forces that are responsible for keeping protons and neutrons together in a stable nucleus.
A
In a nucleus, there is electrostatic force of repulsion between protons
For the nucleus to remain stable, the SNF provides an attractive force between nucleons between 0.5 and 3fm.
Below 0.5fm, the SNF is repulsive to prevent nucleons collapsing into a singularity.
Above 3fm, the SNF is 0 to prevent nucleons in other nuclei being attracted.
14
Q
Explain, using your knowledge of the ranges of the 4 fundamental forces, why heavier elements require more neutrons than protons to be stable.
A
The EMF force acts over an infinite range, while the SNF only acts over distances around 10^-15m.
Each proton in the nucleus will experience EMF repulsion from every other proton in the nucleus. However each nucleon only has the SNF force acting from adjacent nucleons.
Therefore the EMF force of repulsion grows exponentially greater than the SNF force of attraction.
This means more neutrons are required, which are not charged and don’t add EMF rupulsion, to provide the extra attractive force to keep the nucleus stable.
15
Q
Explain why certain types of nuclei undergo alpha decay
A
very large/heavy nuclei
because they are too massive for the SNF to keep them stable.
16
Q
State (and explain, although not required knowledge) which certain types of nuclei undergo beta-minus and beta-plus decay.
A
for beta-minus, neutron rich nuclei
for beta-plus, proton rich nuclei
[NOT REQUIRED KNOWLEDGE] The nuclear shell model says that protons and neutrons exist in energy levels in a nucleus, due to the pauli exclusion principle. If there is an excess of either neutrons of protons, it is energetically favourable for one of the excess to decay into the other and have it fill a lower energy level, where the binding energy is greater.
17
Q
Describe in full detail, why the existence of the electron anti-neutrino was postulated.
A
- emitted beta- particles from beta-minus decay have a range of KE from 0 up to a maximum
- but all beta- particles have energies < theoretical total energy (using E = mc^2)
- to conserve energy, there must be an additional particle that carries away some KE.
- electron antineutrino.
18
Q
State the type of nuclei that undergo gamma decay.
A
nuclei with excess energy, usually after an alpha or beta decay but not always.
19
Q
236/92 U undergoes a series of decays to produce 204/82 Pb. How many alpha decays are involved in this process?
A
8
20
Q
Explain what is meant by the wave-particle duality of light.
A
light exhibits both wave-like and particle-like properties
21
Q
photon
A
a discrete packet/ quantum of EM energy
22
Q
Define the unit of the electron-volt.
A
The energy gained by an electron that passes through a potential difference of 1 volt.
23
Q
Why are electon-positron pairs more likely to be pair produced than any other pair?
A
electrons/positrons have the smallest mass and rest energies
which require the least energy to create
24
Q
Explain why there is a minimum energy in pair production.
A
Conservation of energy means energy before = energy after
photon must supply at least rest energy of particle and anti-particle to create particles
25
Explain why pair production cannot take place if the frequency of the photon is below a certain value.
Photon requires a min energy to create the two particles = 2 x rest energy of one particle
A min energy corresponds to a min frequency
since E = hf, or E d.p to f
26
A gamma photon of energy 1.9 MeV undergoes pair production and creates an electron-positron pair. The energy of the photon is greater than the minimum energy required. Assuming the kinetic energy is split equally, calculate the kinetic energy of one of the particles in MeV.
E before = E after
1.9MeV = 2 x RE + 2 x KE
1.9MeV = 1.022MeV + 2 x KE
0.878MeV = 2 x KE
KE = 0.439MeV
27
In annihilation, a particle and it's corresponding anti-particle meet and destroy each other, producing 2 gamma photons. a) State the equation that gives the minimum energy of a gamma photon produced. b) Explain why the gamma photon is likely to have an energy greater than that predicted by the equation.
a) The equation is:
2RE = 2Emin, so Emin = RE
b) The equation assumes 0 kinetic energy
However the particles must have some kinetic energy in order to collide
28
[NOT REQUIRED KNOWLEDGE] How does a PET scanner work?
patient injected with radioaactive isotope that emits positrons (via beta-plus decay)
the positrons emitted quickly annihilate with electrons
producing 2 gamma photons that can be detected
cancerous cells uptake the radioactive isotope at a faster rate and so produce more photons
so locations of cancerous tumours can be identified.
29
Explain the role of exchange particles for the 4 fundamental forces.
Exchange particles give rise to the forces between particles
by transferring momentum between them
30
Name the 4 fundamental forces and their exchange particles. Give the approximate ranges.
Strong - pions ~10^-15m
EMF - virtual photons infinite
Weak - w bosons ~10^-18m
Gravity - gravitons (theoretical) infinite
31
What type of particles are affected by the weak interaction?
All
32
What types of particles are affected by the strong interaction?
hadrons
33
The weak nuclear force acts over a much shorter distance than the strong force. Explain why this is using your knowledge of exchange particles.
-exchange particles for WNF (W bosons) are much heavier than for SNF (pions)
-particles with greater mass require more energy to create
-particles with greater energies are less stable and have shorter lifespans
-so W bosons travel a shorter distance before decaying.
34
[NOT REQUIRED KNOWLEDGE]
- what are cosmic rays and where do they originate from?
- what are the two main types of particle accelerator and how do they work?
- cosmic rays are high-energy particles or clusters of particles (usually protons or atomic nuclei) which are travelling through space at nearly the speed of light.
- The main two types of particle accelerator are electrostatic and electrodynamic. electrostatic uses a high voltage, while electrodynamic uses a changing electromagnetic field to accelerate particles.
35
Explain why protons are the only stable type of baryons.
Protons are the lightest baryon.
Radioactive decay always causes particles to decay from heavier particles to lighter.
36
Explain in detail, why it is that baryon number is conserved in alpha and beta-minus decay, accounting for all the emitted particles in each case.
in alpha decay, an alpha particle is emitted (2p and 2n), which accounts for 4 baryons.
The daughter nuclei has 2 fewer protons and 2 fewer neutrons, which accounts for -4 baryons.
in beta decay, a neutron turns into a proton, which are both baryons and so does not change baryon number.
The emitted electron anti neutrino and electron are both leptons, not baryons, so do not affect baryon number.
37
Explain why kaons are only produced in pairs, with reference to conservation laws.
Kaons are strange particles
Strange particles are produced via the strong interaction
but strangeness is conserved in the strong interaction
so only pairs of strange particles can be produced, with opposite strangeness.
38
Give the quark composition of K0.
K0 always has +1 strangeness
- d s_bar
39
Explain why kaons are more unstable than pions.
Kaons are heavier than pions, since they have strange quarks that are heavier than down quarks
Particles with greater mass require more energy to create.
Higher-energy particles are more unstable.
40
What do kaons decay into?
pions
41
6 marker : Discuss the nature of hadrons. Key points: 1. classification of hadrons 2. identifying properties, 3. structure of hadrons, 4. stability of free hadrons
-All hadrons are non-fundamental particles which are made up of quarks
- all hadrons experience the SNF and WNF.
- 2 classifications for hadrons include baryons and mesons.
- A baryon is made up of 3 quarks (quark-quark-quark), while a meson is made up of 2 (quark-antiquark pair)
- In general, baryons are more stable than mesons because the meson often annihilates itself after a short period of time
- the proton is the only stable baryon because it is the lightest. free neutrons can decay into protons : n -> p + e- + ve_bar
- although all mesons are unstable, pions are the most stable mesons because they are also the lightest. Kaons will decay into pions.
42
6 marker: Discuss the nature of leptons. Key points: 1. identifying structure
2. interactions with other particles
3. stability and decay of free leptons
4. neutrino and it's undetectability.
5. give an example of lepton decay with an equation.
All leptons are fundamental particles, not made of quarks.
They do not experience the SNF, and interact with other particles via WNF, EMF or gravity only.
Muons are heavier than electrons and so are more unstable. Muons decay into electrons.
The neutrino is a particle with almost zero mass and zero charge.
u- -> vu + e- + ve_bar
43
6 marker: Write an account of the 4 fundamental particle interactions.
Key points:
1. include the names of the interactions
2. identify the groups of particles affected by each
3. identify the exchange particle
4. give examples for 2 of the interactions.
5. include ranges in your answer.
-The Strong nuclear force affects all hadrons only. Mesons interact with baryons via this interaction. Exchange particle is the pion. It's range is approximately 10^-15m.
- The Weak interaction affects all particles, including leptons. It's exchange particle is the W boson, with a range approx = 10^-18m.
- The EMF affects all charged particles only. It's exchange particle is the virtual photon. It has infinite range, since it's exchange particle has no mass.
- Gravity affects all particles with mass. It's exchange particle is the graviton, which is theoretical. It has infinite range.
Examples - Weak: beta minus decay n -> p + e- + ve_bar
Strong: force between nucleons holding the nucleus together
EMF: two electrons
Gravity : gravitational force of Earth on me
44
Give the anti-particles of:
pi-
pi0
K+
K0_bar
pi+
pi0
K-
K0
45
Give 2 characteristics of a strange particle that are different from other non-strange particles.
- they have strange quarks
- they have a longer lifetime than usual
- always produced in pairs
46
Give 2 kaon particles with a strangeness of +1.
K0 and K+
47
Give 2 kaon particles with a strangeness of -1.
K0_bar and K-
48
The following interaction occurs via the strong interaction. Determine the strangeness of X.
K- + p -> K0 + K+ + X
-3
49
The following interaction occurs via the strong interaction.
K- + p -> n + X
X is neutral.
Deterimine the quark composition of X.
sd_bar
50
The equation below shows the decay of a kaon. A pi0 particle decays after 8.4 x 10^-17s.
K+ -> pi0 + e+ + X
1. Name the interaction that causes kaon decay
2. Identify particle X using conservation laws
3. Explain using relevant physics, why it may be difficult to detect any of the products of kaon decay.
1. weak
2. electron neutrino
3. - pi0 decays very quickly since it is a meson made up of a quark-anti quark pair that annihilates quickly
- e+ annihilates quickly with it's corresponding anti-particle, an electron, which is abundant everywhere
- electron neutrino has no charge and almost zero mass, so difficult to detect
51
Explain why you know what interaction governs electron capture, using your knowledge of quarks.
p + e- -> n + ve
weak interaction, as
uud -> udd
u -> d quark character changes
which is only possible in the weak interaction
52
Describe the photoelectric effect.
when photons incident on a metal plate cause the emission of photoelectrons from the surface of the plate
53
Explain why the photoelectric effect occurs only for metals.
There are delocalised electrons in the metal that are not bound to any atom
54
[6 marker] Assuming the wave model of light is correct, predict what should happen in the photoelectric effect.
Experimentally, the wave model of light fails to explain the photoelectric effect.
Explain the observations using the particle model of light instead.
Wave model:
- electrons build up energy over time
- emitted once they have enough energy
- should be a delay between light incident, and photo electrons emitted
Particle model:
- each incident photon has energy directly proportional to it's frequency E = hf
- one photon transfers all it's energy to one electron
- photoelectron emitted instantaneously, if the energy supplied by the photon is greater than the work function
- work function is the minimum energy required to remove an electron from the surface of the metal
- any excess energy of the photon goes into the kinetic energy of the photoelectron
55
Explain, in full detail, why there exists a threshold frequency for the photoelectric effect.
minimum energy is required (work function) to overcome EMF force of attraction between e- and metal ions.
one photon transfers all it's energy to one electron.
a photon's energy is directly proportional to it's frequency (E = hf)
So a minimum energy means a minimum frequency required.
56
Define work function
The minimum energy required to completely remove an electron from the surface of a metal plate
57
Define stopping potential.
The minimum potential difference required to completely stop the photoelectric effect (photoelectron emission)
58
In the photoelectric effect, explain the effect of increasing the intensity of the incident light, on the number of photoelectrons emitted per second.
Increasing the light intensity means
more photons incident per second
one photon transfers all it's energy to one electron
so more photoelectrons emitted per second
59
Explain why increasing the frequency of incident light in the photo electric effect has no effect on the rate of photoelectric emission.
increasing frequency does not increase the number of photon incident per second
each photon transfers all it's energy to one electron
so the number of photoelectrons emitted per second remains the same
60
Use you knowledge of the photoelectric effect to explain why increasing the frequency of incident photons on a metal plate increases the maximum kinetic energy of the emitted photoelectrons.
-Increasing frequency of incident photons increases their energy, since E = hf
-one photon transfers all it's energy to one electron
- Ekmax = hf - work function
-any excess energy is transferred/converted into the KE of the photo electron, so if work function remains constant, Ekmax increases.
61
We can measure the maximum kinetic energy of photo electrons using a photocell. This includes two plates seperated by a gap, a variable power supply connected to the two plates, along with an ammeter in series. Blue light is shone on one of the plates.
The variable power supply is initially set to zero. But there is still a reading on the ammeter.
Explain this phenomenon.
[LONG QUESTION] As the power supply is gradually turned up, the emitter plate becomes more positive.
Explain why this affects the reading on the ammeter.
The power supply is gradually turned higher until the reading on the ammeter is zero.
Explain, using an equation, why this happens and what this voltage is called.
Initially, there is a reading on the ammeter because:
- blue photons have freq greater than threshold freq, and enough energy > work f., causing photoelectrons to be emitted from the surface of the plate
- emitted e-s have enough KE to cross the gap
- so flow of photoelectrons in circuit -> flow of charge
- generating a current
As the power supply is turned up, the reading on the ammeter decreases because:
- the p.d opposes the movement of the photoelectrons across the gap, increasing work required to cross gap
- e-s have a range of KEs, some photoelectrons do not have enough KE to cross the gap
- so less photoelectrons flow in the circuit
- so current decreases
When the power supply is at the point where the reading on the ammeter is zero:
-the work required to cross gap = max KE of electrons
- so no photoelectrons can cross the gap and there is no flow of charge and no current
- it is the stopping potential
62
Give the equation that links stopping potential and maximum kinetic energy of photoelectrons.
Ekmax = eVs
63
Why are energy levels negative?
-energy of a free electron must be 0
-energy must be supplied to an orbital electron to move it to a higher energy level or to free it from the atom
- so relative to the free electron, energy of levels are negative
64
Define excitation.
-when an electron gains energy and transitions to a higher energy level
65
Say we had an atom where the energy required to move an electron from energy level 1 to 2 was 6.7eV. An electron with 26eV of kinetic energy collides with the orbital electron and transfers enough energy to excite it from 1 to 2.
Work out the speed of the electron after the collision.
energy electron after = 26 - 6.7 = 19.3eV = kinetic energy after
19.3 x 1.6x10^-19 = 1/2mv^2
v^2 = (2 x 19.3 x 1.6x10^-19)/ (9.11x10^-31)
v = 2603700m/s
66
Define de-excitation.
when an electron loses energy by emitting a photon and transitions to a lower energy level.
photon has energy exactly equal to the difference in energy between energy levels.
67
When an electron deexcites, is there more than one way this can occur?
yes! cascading: the process by which excited electrons return to the ground state via different de-excitation pathways.
68
Define ionisation.
When an electron gains enough energy to be removed from an atom: i.e becomes a free electron not bound to the atom.
69
Define ionisation energy
The minimum energy required to completely remove an electron in the ground state
70
An electron is incident on a hydrogen atom. As a result an electron in the ground state is excited to a higher energy level. Explain why the electron in the ground state is excited to a higher energy level.
high-energy free electron collides with orbital electron and transfers KE
energy transferred is exactly equal to the difference between energy levels
71
State what is meant by "ground state" with respect to electrons and atoms.
An electron in the lowest energy level, n = 1, closest to the nucleus
72
Explain the difference between ionisation and excitation.
in ionisation, the electron gains enough energy to become a free electron: no longer bound to the atom
while in excitation, the electron gains exactly enough energy to transition to a higher energy level, but it is still bound to the atom.
73
An atom can also become excited by the absorption of photons. Explain why only photons of certain frequencies cause excitation in a particular atom.
-electrons exist in discrete energy levels
-1 photon transfers all it's energy to 1 electron
-for excitation, electrons requires the photon has an specific energy equal to the energy difference between energy levels
-photons of specific energies correspond to photons of specific frequencies via E = hf
74
Describe and explain in full detail, how a fluorescent tube works.
- high p.d across tube causes high energy electrons to flow through tube
- high energy free electrons collide with orbital electrons in mercury atoms, transferring KE to the electrons exactly equal to the energy difference between levels, exciting them to higher energy levels
- which are unstable and deexcite, emitting photons of UV light
- UV photons are absorbed by orbital electrons in atoms of phosphor coating, exciting them
- excited electrons deexcite via different pathways in a process called cascading, causing photons of lower energies and frequencies (E=hf) to be emitted in the visible spectrum.
75
The mercury atoms in a fluorescent tube are excited, and then emit photons in the ultraviolet region of the electromagnetic spectrum.
Explain how the mercury atoms become excited.
Explain how these excited mercury atoms then emit photons.
Explain how the ultraviolet photons in the tube are converted into photons in the visible part of the electromagnetic spectrum.
- high energy free electrons collide with orbital electrons in mercury atoms, transferring KE equal to exactly the energy difference between levels, exciting it to a higher energy level
- excited electrons are unstable and deexcite to lower energy levels, emitting photons with energy equal to the difference in energy between levels.
- ultraviolet photons are absorbed by atoms of the phosphor coating, exciting orbital electrons to higher energy levels
- excited atoms deexcite and lose energy, emitting photons. orbital electrons deexcite to lower energy levels, moving to ground state via other energy levels
- resulting in a range of lower photon energies/frequencies.
76
Why are only certain wavelengths seen in a light emission spectra?
- electrons only exist in discrete energy levels
- excited electrons in gas deexcite to lower energy levels
- emitting specific energy photons with exact energy difference between levels
- which correspond to specific wavelengths as E = hc/lambda.
77
[6 marker] Explain how emission line spectra are produced. In your answer you should describe:
- how the collisions of charged particles with gas atoms can cause the atoms to emit photons
- how spectral lines are explained by the concept of discrete energy levels
- high energy free electrons collide with orbital electrons in gas atoms
- transferring KE exactly equal to the energy difference between energy levels
- exciting them to higher energy levels
- excited electrons are unstable and de-excite to lower energy levels, emitting a photon of specific energy equal to the energy difference between levels
- electrons exist in discrete energy levels, so discrete photon energies emitted
- each spectral line corresponds to a specific energy transition in the energy levels
- larger energy difference -> higher energy photon -> lower wavelength as E = hc/lambda.
78
Explain how line absorption spectra are formed.
-white light is shone through a cold gas
-orbital electrons absorb photons with exact energy equal to the energy difference between energy levels
-electrons exist in discrete energy levels, so only discrete photon energies absorbed
-discrete photons energies correspond to discrete photon wavelengths by E = hc/lambda
79
Give the de broglie's relationship for the wavelength of a particle.
lambda = h/p
80
Define diffraction.
The spreading out of waves as they pass through an apeture or gap.
81
When is the diffraction effect at a maximum, in relation to gap size and particle wavelength?
The maximum diffraction effect occurs when the gap size is very close in size to the particle's wavelength.
82
In electron diffraction, electrons are accelerated by a potential difference to a high speed, after which they pass through a small gap. A phosphor screen is placed at the far end, after the electrons pass through the gap. Small flashes of light are observed at the screen.
Give 2 examples of particle like behaviour in this experiment.
Explain why small flashes of light are observed at the screen.
2 examples of particle like behaviour here are:
- electrons being accelerated by a p.d
- fluorescence at the screen
The small flashes of light is caused by fluorescence: -high-energy free electrons collide with orbital electrons of atoms in the screen, exciting them by transferring KE.
- excited electrons lose energy and dexcite to lower energy levels, emitting photons
83
Describe how line absorption spectra are formed, with reference to energy levels. [4 marks]
-discrete energy levels
-white light, cold gas
-excitation of e-s, gain E by absorption of one photon
-E of photon absorbed = difference between levels
-but E = hc/lambda, specific wavelengths of photons absorbed
-each transition associated with absorption of a unique wl
84
Describe how a line emission spectra is formed, in relation to changes in energy. [4 marks]
-discrete energy levels
-excitation, gain E by collision of free e-
-deexcitation, loss E by emission of a photon
-E of photon emitted = difference between energy levels
-but E = hc/lambda, specific wavelengths of photon emitted
- each transition associated with emission of a unique wl
85
86
Give examples of light behaving as a wave, and as a particle.
Wave: - diffraction (Young's double slit)
- refraction (change in direction of propagation of a wave when passing from one medium to another)
Particle: - photoelectric effect
- excitation/de-excitation
87
88
In an arrangement used to investigate the photoelectric effect, a metal plate and a point electrode with a gap inbetween are connected to a circuit with a variable power supply, connected in series to an ammeter.
As the variable power supply is turned up, keeping all other factors unchanged, the ammeter reading rises to a maximum. Explain why there is a maximum current reached.
- the p.d causes electrons to be attracted to the electrode
- above a certain p.d, all photoelectrons that are emitted reach the electrode
- but the rate of photoelectron emission is limited by the intensity of light
- so flow of photoelectrons is limited by intensity
- current is limited.
89
90
An isolated metal plate is given a negative charge. EM radiation is incident on the plate. The plate loses it's charge due to the photoelectric effect.
Discuss how the rate of loss of charge from the plate depends on both the frequency and the intensity of the incident radiation.
In your answer, explain why:
- the plate loses it's charge
- the photoelectric effect occurs only for frequencies greater than a particular value
- the rate of loss of charge increases with intensity for radiation above that particular value of frequency
Loss of charge?:
- in the photoelectric effect, incident EM radiation on the surface of a metal plate causes emission of photoelectrons from the plate. Since electrons are negatively charged, and the plate is losing e-s, the plate is losing it's negative charge.
Threshold frequency:
- for e-s to be emitted, the delocalised electrons must overcome the EMF force of attraction between e-s and positive metal ions in the metal plate.
- one photon transfers all it's energy to one e-
- so the photon must supply a minimum energy = work function (min energy to completely remove an electron from the surface of a metal plate)
- a minimum energy corresponds to a minimum frequency: E = hf.
intensity vs frequency on rate:
- as intensity is increased, the number of photons incident on the plate per second increases
- one photon transfers all it's energy to one e- (1-to-1 correspondence)
- so an increased number of photoelectrons emitted per second
- increased rate of loss of charge
Frequency
- as frequency is increased, the number of photon incident per second does not change, so the rate of loss of charge stays the same.
- however, the energy of the incident photons increases, so the maximum KE of the photoelectrons increases.
91
92
Give the rough wavelengths in nm, of the EM spectrum from gamma to Visible light.
0.1nm Gamma (x200)
20nm XRAY (x20)
400nm UV (x2)
800nm Vis
