Exam 2 Review

Question 1

When Roentgen discovered x-rays, he was working with a ________ tube.

Answer 1

Crookes

When Wilhelm Roentgen discovered x-rays in 1895, he was experimenting with a Crookes tube, which is a type of early vacuum tube used to study electrical discharges in gases.

Question 2

The “quality” of an x-ray beam refers to its_____________

Answer 2

penetrating power.

Question 3

Which type of interaction best illustrates the conversion of energy into mass?

Answer 3

Pair production

In pair production, an incoming photon interacts with the electromagnetic field of an atom’s nucleus.

The photon disappears and its energy is used to create an electron and a positron.

The threshold energy for pair production to occur is 1.022 MeV. This is because 1.022 MeV is equivalent to the combined rest mass energy of an electron and a positron.

The probability of pair production occurring increases with the atomic number of the material.

Pair production is only important at energies above 10 MeV.

This interaction is a direct example of Einstein’s famous equation, E=mc², which states that energy and mass are equivalent and can be converted into one another. In pair production, the energy of the photon is converted into the mass of the electron and positron.

Question 4

Bremsstrahlung radiation is produced when

Answer 4

an electron is decelerated and loses part of its kinetic energy.

Question 5

The x-ray emission in an x-ray tube is considered to be

Answer 5

heterogenous

The x-ray emission in an x-ray tube is considered to be heterogeneous because the x-rays produced have a range of energies. They are not monoenergetic, meaning they do not have a single energy level but instead a spectrum of different energy levels.

Question 6

High voltage applied across an x-ray tube will result in a beam with

Answer 6

high penetration & short wavelength

High penetration: The higher the voltage (kVp), the more energy the x-rays have, allowing them to penetrate more deeply into tissues.

Short wavelength: Higher energy x-rays have shorter wavelengths, which are associated with more penetrating power.

Low frequency would be INCORRECT because higher energy x-rays are associated with higher frequency, not lower.

Question 7

In x-ray production, the side of the tube containing the target is the ____________. The target in diagnostic x-ray equipment is the type of target called a ____________ target.

Answer 7

anode; reflection

Question 8

The minimum wavelength in an x-ray beam depends on

Answer 8

the peak kilovoltage

The minimum wavelength in an x-ray beam is determined by the peak kilovoltage (kVp) applied to the x-ray tube. Higher kVp results in higher energy x-rays, which have shorter wavelengths. Therefore, increasing the kVp reduces the minimum wavelength of the x-rays produced.

Tube current, source-image distance, and HVL (half-value layer) affect other aspects of the x-ray beam but not the minimum wavelength.

Question 9

Conversion of a positron – electron pair to a pair of photons is called:

Answer 9

Annihilation reaction

Pair annihilation is the opposite of pair production.

In pair annihilation, a positron (e+), which is the antiparticle of an electron, combines with an electron.
When this happens, both particles are annihilated.
This annihilation produces two photons that travel in opposite directions.

Question 10

In order to decrease penumbra, one would

Answer 10

Penumbra is the blurring of image edges in x-rays.

To decrease penumbra, you should:
Increase SFD (move the source farther from the film).

Decrease source size (use a smaller focal spot).

Decrease OFD (bring the object closer to the film).

Question 11

Radioactivity was discovered by ___________

Answer 11

Henri Becquerel

Question 12

The emulsion on an x-ray film consists of 95% of what?

Answer 12

Silver bromide

The emulsion on an x-ray film consists of approximately 95% silver bromide crystals. These crystals are light-sensitive and play a key role in creating the image when exposed to x-rays. The remaining 5% is usually silver iodide.

Question 13

When an electron is removed from an atom, the atom is said to be:

Answer 13

Ionized

Question 14

A positron loses energy passing through matter and eventually combines with an electron resulting in ___________

a) two photons of equal energy emitted in opposite directions

b) two particles emitted in opposite directions

c) a photon with energy equal to the rest mass of both articles

d) a particle with equal mass but no charge

Answer 14

The answer is two photons of equal energy emitted in opposite directions.

When a positron encounters an electron, they undergo a process called pair annihilation. This process results in the complete conversion of their mass into energy. Since the positron is the antiparticle of an electron, they have equal but opposite charges and equal mass. During pair annihilation, this mass is converted into energy in the form of two photons.

To conserve momentum, these two photons are emitted in opposite directions. Additionally, to conserve energy, the total energy of the two photons must equal the total energy (rest mass energy plus kinetic energy) of the electron-positron pair before annihilation.

Let’s look at the other answer choices and why they are incorrect:

two particles emitted in opposite directions: Pair annihilation results in the complete conversion of the particles into energy, not the emission of other particles.
a photon with energy equal to the rest mass of both particles: While the total energy of the photons equals the total energy of the electron-positron pair, this energy is distributed into two photons, not one.
a particle with equal mass but no charge: Pair annihilation results in the destruction of the electron and positron, not the creation of a new particle.

Question 15

Regarding x-ray tubes, all of the following are true:

Answer 15

The filament emits electrons by thermionic emission.

The kVp is the peak voltage applied between the anode and cathode.

When electrons strike the target, characteristic x-rays and Bremsstrahlung are emitted.

The target is angled and rotated to increase its heat capacity.

Electrons travel from the cathode to the anode.

Question 16

A Compton-scattered electron:

a) is absorbed within a few microns of the site of the original Compton interaction.

b)annihilates another electron.

c)causes pair production.

d)engages in the process of photodisintegration.

Answer 16

is absorbed within a few microns of the site of the original Compton interaction.

The Compton effect, also known as Compton scattering, is an interaction between an incoming photon and a loosely bound outer shell electron, which is treated as a free electron.
During this interaction, the photon imparts some of its energy to the electron, ejecting it from the atom. This ejected electron is known as a Compton electron.
The photon, now carrying reduced energy, changes direction and is known as a scattered photon.
The Compton electron, being a charged particle, will quickly interact with nearby atoms and deposit its energy within a short distance, typically a few microns, from the site of the Compton interaction. Let’s look at why the other options are not correct:

(b) annihilates another electron: Annihilation occurs when a positron encounters an electron, resulting in the conversion of their combined mass into energy, typically emitted as two photons. The Compton electron is a negatively charged electron and won’t annihilate with another electron.
(c) causes pair production: Pair production requires an incoming photon with energy above a threshold of 1.022 MeV to interact with the electromagnetic field of a nucleus. While the Compton electron carries kinetic energy, it is not a photon and cannot directly initiate pair production.
(d) engages in the process of photodisintegration: Photodisintegration involves a high-energy photon being absorbed by a nucleus, making it unstable. This process typically occurs at energies above 10 MeV. The Compton electron does not possess the necessary energy or properties to cause photodisintegration.

Question 17

Following a photon interaction with matter, a photon is detected. It could be any of the following except:

a)a scattered photon following photoelectric interaction.

b)an annihilation photon following pair production.

c)a scattered photon following Compton interaction.

d)a characteristic x-ray following photoelectric interaction.

Answer 17

a)a scattered photon following photoelectric interaction.

The photoelectric effect is an interaction where an incoming photon interacts with a tightly bound inner orbit electron and transfers all of its energy to the electron.
The electron is ejected from the atom with kinetic energy equal to the energy of the incident photon minus the electron’s binding energy. This ejected electron is called a photoelectron.

The atom is left with a positive charge, and outer orbit electrons drop to inner orbits to fill the vacancy.
The energy differences from the electrons dropping to lower energy levels are released as photons called characteristic x-rays.
Since all the energy from the incident photon is absorbed in the interaction, there is no scattered photon.

Here’s why the other options are possible:

(b) an annihilation photon following pair production: Pair production results in the creation of a positron and an electron. When a positron comes to rest, it combines with an electron, and both are annihilated, producing two annihilation photons that travel in opposite directions.
(c) a scattered photon following Compton interaction: A relatively high-energy photon interacts with a loosely bound electron in the atom’s outer shell. Some of the energy from the photon knocks the electron out of its orbit. The remaining energy emerges as a new scattered photon with reduced energy and a change in direction.
(d) a characteristic x-ray following photoelectric interaction: As explained above, when an electron is ejected during the photoelectric effect, characteristic x-rays are emitted as outer shell electrons drop down to fill the vacancy.
Therefore, the only option that is not possible is a scattered photon following photoelectric interaction.

Question 18

“ALARA” stands for

Answer 18

As Low As Reasonably Achievable

Question 19

Which subatomic particle can cause the most biologic damage when used in radiation therapy?

Proton
Photon
Electron
No answer text provided.

Answer 19

Proton

Question 20

Which of the following statements is NOT true of ionizing radiation?

a) Ionized atoms have a negative charge

b)Its particles have enough energy to cause an electron to become free of its atom

c)It has enough energy to overcome the binding energy within an atom

Answer 20

a) Ionized atoms have a negative charge

Question 21

An atom that has gained energy is said to be in what state?

Answer 21

Excited

Question 22

Characteristic radiation is so named because:

Answer 22

The energy of the photons emitted is dependent on individual elements

Question 23

Which of the following is true about bremsstrahlung radiation?

Answer 23

A beam electron looses energy and a photon is emitted

Question 24

What does the stopping power depend on?

Answer 24

The beam energy and the material

Question 25

What is NOT true of the Compton effect?

a)It is seen in beams with energies between 100 keV and 20 MeV

b)It is the dominant interaction in tissue during radiation therapy

c)The probability of interaction depends on the material density

d)It involves the beam electrons interacting with the atom nucleus

Answer 25

d)It involves the beam electrons interacting with the atom nucleus

because in Compton effect the photon interacts with a loosely bound electron in the atoms (outer orbit) no where near the nucleus.

Question 26

The most important clinical aspect of beam hardening is?

Answer 26

It reduces the number of photons, but increases the average energy of the beam

Question 27

X-ray photon beam attenuation is influenced by

Answer 27

1. tissue type.
2. subject thickness.
3. photon quality.

Question 28

The line focus principle expresses the relationship between

Answer 28

the actual and the effective focal spot

Question 29

How is SID ( Source-to-Image Distance) related to exposure rate and radiographic density?

Answer 29

As SID increases, exposure rate decreases and radiographic density decreases.

Exposure rate decreases: The farther the x-rays travel, the more they spread out, and the fewer x-rays hit the image receptor. This means the exposure rate (the intensity of the radiation reaching the receptor) decreases.

Radiographic density decreases: As fewer x-rays hit the image receptor, the image becomes lighter, which results in a decrease in radiographic density. Radiographic density refers to the degree of blackness on a developed x-ray film. More x-rays hitting the receptor result in a darker image, and fewer x-rays result in a lighter image.

Underexposure (due to low exposure rate) can result in an image that is too light or lacks enough detail. In this case, you might need to adjust the exposure settings (like increasing the milliampere-seconds or mAs) to compensate for the lower density caused by the increased SID.

Moving the x-ray source further away from the film does decrease penumbra, which is the blurring around the edges of structures in the image. This improves image sharpness.

In practical terms, increased SID is often used to improve image sharpness and reduce penumbra, especially in specific radiographic techniques like chest X-rays. However, you must adjust other factors like mAs to ensure the image has enough exposure and proper density.

Question 30

Which of the following affect(s) both the quantity and the quality of the primary beam?

1. Half-value layer (HVL)
2. kVp
3. mA

Answer 30

1 and 2 only

1. Half-value layer (HVL)
2. kVp

Not mA because that only affects the quantity of x-rays.

Question 31

The reduction in x-ray photon intensity as the photon passes through material is termed

Answer 31

Attenuation

Question 32

For SAD treatments, the field size is determined at the:

Answer 32

Isocenter

because SAD means source to axis distance, which is 100cm at isocenter.

Question 33

X-rays were discovered by ____________________ in the year _________.

Answer 33

Wilhelm Roentgen; 1895

Question 34

The process when two high-energy photons are emitted 180 degrees away from each other is termed:

Answer 34

Annihilation process

Question 35

The inverse square law states that if you double the distance, the intensity

of x-rays __________its original value.

Answer 35

Decreases to 1/4

Question 36

The x-ray interaction that is predominant in the therapeutic energy range is

Answer 36

Compton

Question 37

The interactions that occur in the x-ray tube are

1. photoelectric absorption
2. Bremsstrahlung interaction
3. characteristic absorption

Group of answer choices

2 and 3 only

1 and 2 only

1, 2, and 3

2 only

Answer 37

2 & 3

Bremsstrahlung & characteristic

Question 38

When more current is applied to an x-ray tube cathode,
Group of answer choices:

a) the quantity of the x-ray beam is increased.

b) the quality of the x-ray beam is increased.

c) the quantity and quality of the x-ray beam are increased.

d)no effect will be noticed.

Answer 38

a) the quantity of the x-ray beam is increased.

Question 39

Radium was discovered by _______________________.
Group of answer choices

Pierre and Marie Curie

Wilhelm Roentgen and Henri Becquerel

Henri Coutard and Claude Regaud

Pierre Curie and Wilhelm Roentgen

Answer 39

Pierre and Marie Curie

Explanation: Radium was discovered by Pierre and Marie Curie in 1898. They isolated radium from uranium ore and noted its intense radioactivity, which became a foundational discovery in the field of radiation therapy and nuclear physics.

Question 40

The reason for the differential contrast between bone and soft tissue in a diagnostic radiograph is due primarily to
Group of answer choices

photoelectric effect.

Compton interactions.

pair production.

coherent scatter.

Answer 40

Photoelectric effect

In the photoelectric effect, an x-ray photon collides with an atom and transfers all its energy to one of the atom’s inner-shell electrons. Here’s a breakdown:

Absorption: The photon gives up all its energy to an inner-shell electron (usually a K-shell electron).

Ejection: This energy boost ejects the electron from the atom, creating an ion.

Electron vacancy: The vacancy left in the inner shell causes an electron from a higher shell to drop down to fill it.

Characteristic x-rays: As the higher-shell electron moves down, it releases energy in the form of characteristic x-rays, but these are usually absorbed by surrounding tissues.

Key Point: The photoelectric effect is more likely in materials with high atomic numbers (like bone) and lower-energy x-rays, which is why it creates contrast between bones and soft tissue.

Question 41

The type of interaction in which a 100 keV photon is scattered with half of its initial energy and an electron is emitted with the remaining energy would be:
Group of answer choices

Compton

Classical scatter

Pair production

Photoelectric

Answer 41

Compton

Explanation: In a Compton interaction (also called Compton scattering), an x-ray photon collides with an outer-shell electron. The photon transfers part of its energy to the electron, ejecting it from the atom and causing ionization. The photon then continues in a different direction with reduced energy.

In this case, the 100 keV photon is scattered with half of its initial energy (50 keV), while the remaining energy (also 50 keV) is transferred to the ejected electron. This characteristic energy distribution and scattering make it a typical example of Compton scattering.

Question 42

Pair production can occur for which of the following photon energies?

1. 1.02 keV
2. 0.51 MeV
3. 1.02 MeV
4. 2.51 MeV

Group of answer choices

3 and 4

4 only

1, 2, 3, and 4

2, 3, and 4

Answer 42

3 and 4

Explanation: Pair production requires a photon with energy at least 1.02 MeV because this is the minimum energy needed to create an electron-positron pair (each with a rest mass energy of 0.51 MeV).

1.02 MeV (option 3) is the threshold energy for pair production to occur.

2.51 MeV (option 4) is also sufficient for pair production, as it exceeds the threshold.

Energies lower than 1.02 MeV (such as options 1 and 2) do not have enough energy to produce an electron-positron pair.

Question 43

Which of the following places human beings in closer contact with extraterrestrial radiation?
Group of answer choices

a flight on a commercial aircraft

posteroanterior and lateral chest radiographs in a doctor’s office

deep-sea diving

a visit to a nuclear power plant

Answer 43

a flight on a commercial aircraft

Explanation: During a flight on a commercial aircraft, passengers are exposed to higher levels of cosmic (extraterrestrial) radiation. At high altitudes, the Earth’s atmosphere provides less protection from cosmic rays, increasing exposure to this type of radiation. The other options—chest radiographs, deep-sea diving, and visiting a nuclear power plant—do not expose individuals to increased extraterrestrial radiation.

Question 44

In which of the following x-ray interactions with matter is the energy of the incident photon completely absorbed?
Group of answer choices

Photoelectric

Compton

Incoherent

Rayleigh

Answer 44

Photoelectric

Explanation: In the photoelectric interaction, the energy of the incident x-ray photon is completely absorbed by an inner-shell electron in an atom. This absorption causes the electron to be ejected from the atom, resulting in ionization. The other interactions (Compton, Incoherent, and Rayleigh) involve scattering, where the photon’s energy is only partially absorbed, and the photon continues with reduced energy or in a different direction.

Question 45

A reduction in the number of primary photons in the x-ray beam through absorption and scatter as the beam passes through the object in its path defines what?
Group of answer choices

Attenuation

Annihilation

Photodisintegration

Radiographic fog

Answer 45

Attenuation

Annihilation is related to the interaction between matter and antimatter.

Photodisintegration involves high-energy photons interacting with an atomic nucleus, causing particle emission(neutrons), which typically occurs at energies much higher than those used in diagnostic imaging.

Radiographic fog is unrelated to photon reduction; it refers to unwanted exposure or scatter that reduces image contrast.

Question 46

A positron is considered
Group of answer choices

a form of antimatter.

a modified proton.

a form of small angle scatter.

a byproduct of the photoelectric interaction.

Answer 46

A form of antimatter

Question 47

Which of the following statements regarding photoelectric interactions is false?
Group of answer choices

the probability increases rapidly with increasing energy

they are mainly responsible for differential attenuation in diagnostic radiographs

the incident photon is absorbed

bound electrons are involved

Answer 47

the probability increases rapidly with increasing energy

Explanation: This statement is false because, in photoelectric interactions, the probability actually decreases with increasing photon energy. Photoelectric interactions are more likely to occur at lower photon energies and with materials of higher atomic number (like bone), which is why they play a significant role in creating differential attenuation and contrast in diagnostic radiographs. The other statements are correct:

Photoelectric interactions are responsible for differential attenuation.

The incident photon is absorbed entirely.

Bound (inner-shell) electrons are involved in the interaction.

Question 48

In pair production, which of the following is true?
Group of answer choices

annihilation photons have an energy of 0.51 MeV

the incident photon is scattered with reduced energy

a pair of orbital electrons are ejected from the atom

two positrons are emitted at 180 degrees

it cannot occur if the photon energy is above 1.02 MeV

Answer 48

annihilation photons have an energy of 0.51 MeV

The other statements are incorrect:

The incident photon is completely absorbed (not scattered with reduced energy).

Only one electron and one positron (not two positrons or two electrons) are produced.

Pair production requires at least 1.02 MeV, but it can occur at energies above this threshold.

Question 49

Compton scatter:
Group of answer choices

decreases in probability as photon energy increases

takes place with an inner shell, bound electron

is an interaction with the nucleus

is more likely at energies just below the K edge

is responsible for the efficiency of lead as a diagnostic room shielding material

Answer 49

decreases in probability as photon energy increases

The other options are incorrect:

Inner shell electrons are primarily involved in the photoelectric effect, not Compton scatter.

Compton scatter does not involve the nucleus; it interacts with electrons.

The K edge is related to photoelectric interactions rather than Compton scattering.

Lead shielding is effective mainly due to its high atomic number and effectiveness against photoelectric interactions, rather than Compton scatter alone.

Question 50

The passage of x-ray photons through an object without interaction is called
Group of answer choices

transmission

absorption

attenuation

scattering

Answer 50

Transmission

Transmission refers to the passage of x-ray photons through an object without any interaction (such as absorption or scattering). In transmission, the photons continue in their original path and contribute to the formation of the image.

Question 51

When electrons strike the anode of an x-ray unit, all of the following can occur except
Group of answer choices

electron capture

characteristic x-ray production

electron deceleration

ionization

Bremsstrahlung production

Answer 51

electron capture

Explanation: When electrons strike the anode in an x-ray unit, electron capture does not occur. Electron capture is a process that happens in certain types of atomic nuclei, not in an x-ray tube.

The other processes listed are possible outcomes in an x-ray tube

Question 52

A 3 MeV photon interacts by the pair production process. What is the combined initial kinetic energy of the positron and electron pair?
Group of answer choices

1.98 MeV

1.02 MeV

2.49 MeV

3.00 MeV

4.02 MeV

Answer 52

1.98 MeV

Explanation: In pair production, a photon with energy at least 1.02 MeV interacts with the nucleus to produce an electron and a positron pair. The energy of the incoming photon is divided as follows:

1.02 MeV is used to create the electron and positron pair (0.51 MeV each, which is their rest mass energy).

The remaining energy goes into the kinetic energy of the electron and positron.

So, for a 3 MeV photon:

3 MeV - 1.02 MeV = 1.98 MeV, which is the combined initial kinetic energy of the electron and positron.

Question 53

When a radionuclide decays, radiation is emitted from the:
Group of answer choices

the nucleus

outer shell electrons

inner shell electrons

all of these

Answer 53

the nucleus

Explanation: When a radionuclide decays, the radiation is emitted from the nucleus. Radioactive decay involves changes within the nucleus, such as the emission of alpha particles, beta particles, or gamma rays, as the nucleus seeks a more stable state.

Outer and inner shell electrons are not involved in radioactive decay. Instead, they are associated with other types of interactions, like ionization or characteristic x-ray production in an x-ray tube.

Question 54

A positron loses energy passing through matter and eventually combines with an electron resulting in ____________.
Group of answer choices

two photons of equal energy emitted in opposite directions

two particles emitted in opposite directions

a photon with energy equal to the rest mass of both articles

a particle with equal mass but no charge

Answer 54

two photons of equal energy emitted in opposite directions

Explanation: When a positron loses energy as it passes through matter, it eventually encounters an electron. The positron and electron annihilate each other, resulting in the production of two photons. These photons each have equal energy (0.51 MeV, which is the rest mass energy of an electron or positron) and are emitted in opposite directions (180 degrees apart) to conserve momentum.

Question 55

90% Isodose Line:

The formula is Energy (E) ÷ 4 = 90% isodose line.

This means: If you take the energy of the beam and divide it by 4, you find out how deep the radiation goes to deliver 90% of the dose.

Example: If the energy is 12 MeV, you would do:

Answer 55

12 MeV ÷ 4 = 3 cm. This tells you that at 3 cm deep, the tissue is receiving 90% of the radiation.

Question 56

80% Isodose Line:

The formula is Energy (E) ÷ 3 = 80% isodose line.

This means: Dividing the energy by 3 shows where the radiation reaches 80% of the dose.

Example: Using 12 MeV again:

Answer 56

12 MeV ÷ 3 = 4 cm. At 4 cm deep, the tissue receives 80% of the radiation.

Question 57

50% Isodose Line:

The formula is Energy (E) ÷ 2.33 = 50% isodose line.

This means: If you divide the energy by 2.33, it tells you where 50% of the dose reaches.

Example: With 12 MeV:

Answer 57

12 MeV ÷ 2.33 ≈ 5.15 cm. At approximately 5.15 cm deep, the tissue gets 50% of the radiation.

Question 58

10% Isodose Line:

The formula is Energy (E) ÷ 2 = 10% isodose line.

This means: Dividing the energy by 2 shows how deep the radiation reaches to give 10% of the dose.

Example: Again using 12 MeV:

Answer 58

12 MeV ÷ 2 = 6 cm. At 6 cm deep, the tissue receives 10% of the radiation.

Question 59

Problem:
What is the energy of an electron beam that reaches the 50 isodose line at 2.58 cm of depth?

Solution:
Energy (E) ÷ 2.33 = 50% isodose line
E ÷ 2.33 = 50% isodose line or depth in cm
or;

Depth x 2.33 = E (formula rearranged)

Answer 59

2.58 x 2.33 = 6.01

Answer:
6 MeV

Question 60

What is the Radiation Quality Factors (QF) for

X and Gamma Rays:

Answer 60

1

Question 61

What is the Radiation Quality Factors (QF) for

Fission Fragments & Heavy Nuclei:

Answer 61

20

Question 62

What is the Radiation Quality Factors (QF) for

Electrons:

Answer 62

1

Question 63

What is the Radiation Quality Factors (QF) for

Alpha Particles & Fast Neutrons:

Answer 63

20

Question 64

What is the Radiation Quality Factors (QF) for

Beta Particles, Positrons, & Muons

Answer 64

1

Question 65

What is the Radiation Quality Factors (QF) for

Neutrons and Protons:

Answer 65

10

Question 66

What is the Radiation Quality Factors (QF) for

Protons with energy > 2MeV

Answer 66

2

Question 67

What is the Radiation Quality Factors (QF) for

Thermal Neutrons:

Answer 67

5

Question 68

Isodose curves are lines on a graph that connect points where the amount of radiation is the same. For example, if one line shows 50% of the radiation, every point on that line gets the same amount of radiation.

Answer 68

Curves:
The curves on the graph represent lines of equal radiation dose. For example, one line might represent a 100% dose, another a 80% dose, and so on.

Labels:
Each curve is usually labeled with a percentage. This tells you how much radiation is delivered at that point. Higher percentages indicate more radiation.

Question 69

X-Axis (Horizontal):
This usually represents distance from the radiation source or the area being treated (like how deep the radiation is going into the body).

Y-Axis (Vertical):
This often represents the percentage of dose or the amount of radiation received at different depths. It shows how much radiation is delivered at each distance.

Answer 69

Question 70

The slides show photos or diagrams of isodose curves. These pictures often display how the curves look in different scenarios, like when different energy levels are used.

Answer 70

Question 71

Close Lines:
If the lines are close together, it means there is a steep gradient of radiation dose, meaning the dose changes quickly over a small distance. This is often seen near the radiation source.

Wider Lines:
If the lines are far apart, the dose changes gradually. This usually means that the radiation is spreading out and less is being delivered to that area.

Answer 71

Target Area:
Look at where the curves indicate the highest doses. This area represents where the tumor is being targeted for treatment.

Safe Zones:
The areas outside the highest dose lines (e.g., the 80% or 90% lines) represent the healthy tissues that should receive less radiation. Understanding where these curves lie helps in planning to minimize damage to healthy cells.

Question 72

Old units

Answer 72

Rad = 100 Erg/g

Rem = QF x Rad

1 Roentgen (R) = 2.58 x 10^-4 C/Kg

1 Curie (Ci) = 3.7x10^10 Bq or dps

Question 73

New Units

Answer 73

Gray (Gy) = 1 J/Kg

Sievert (Sv) = QF x Gy

1 Coulomb (C)/Kg = 3.88 x10^-3 R

1 Becquerel (Bq) = 2.7 x 10^-11Ci or dps

Question 74

1 Gy = how many Rads

Answer 74

100 Rads

Question 75

1 cGy = how many Rads

Answer 75

1 Rad

Question 76

1 Rad = how many Gy

Answer 76

0.01 Gy or 1 cGy

Question 77

1 Sv = how many Rems

Answer 77

100 Rem

Question 78

1 mSv = how many Rems

Answer 78

0.1 Rem or 100 mRem

Question 79

1 Rem = how many cSv

Answer 79

1 cSv or 0.01 Sv

Question 80

1 Rem = how many Sv

Answer 80

0.01 Sv or 1cSv

Question 81

Rem x 10 =

Answer 81

mSv

Question 82

mSv ÷ 10 =

Answer 82

Rem

Question 83

Problem:
18 Gy equals how many Rad?

Answer 83

Solution:
1 Gy = 100 Rad
18 Gy x 100 Rads per Gy = 1800 Rad

Answer:
1800 Rad

Question 84

Problem:
850 mSv equals how many Rem?

Answer 84

Solution:
1 mSv = 0.1 Rem
850 mSv x 0.1 Rem per mSv = 85 mSv

Answer:
85 mSv

Question 85

Problem:
99 Rem equals how many Sv?

Answer 85

Solution:
1 Rem = 0.01 Sv
99 Rem x 0.01 Sv per Rem = 0.99 Sv

Answer:
.99 Sv

Question 86

Dose equivalent is the amount of standard x-ray radiation that has the same biological effect as another, non-standard, type of radiation.

Dose equivalent = absorbed dose x QF

Dose equivalent is in…

Answer 86

Dose equivalent is in Sieverts (Sv) or Radiation Equivalent to Man (REM).

Rem = Rad x QF

Sv = dose (Gy) x QF

Question 87

Question 88

Problem:
2.5 Rads of neutrons equals how many Rems and Sieverts?

Answer 88

Question 89

Problem:
45 Rads of alpha particles equals how many Rems and Sieverts?

Answer 89

Question 90

Problem:
8.5 Rads of gamma rays equals how many Rems and Sieverts?

Answer 90