Question

In: Physics

a) A tumour located at a depth of 10 cm is treated using a non-isocentic technique...

a) A tumour located at a depth of 10 cm is treated using a non-isocentic technique with a Co-60 machine, to a dose of 1.8 Gy, with a single 5 cm x 5 cm beam. Compute the maximum dose received anywhere in the patient's anatomy. Refer to the attached table for PDD data. (5 pts) b) At what depth does the maximum dose occur? Explain. (5 pts) b) Compute the dose at a depth of 5 cm along the beam axis. (5 pts) c) Would this be a good treatment strategy? Explain. (5 pts)

Solutions

Expert Solution

a)Attached Table:
Field Size in (cm2)
5 x 5
10 x 10
20 x 20
Depth in cm
0.5
100
100
100
5.01
76.72
80.40
83.2
10.0
53.6
58.74
63.38

this a table of PDD data.

b)Cobalt 60 is produced in a nuclear reactor by bombarding 59Co with neutrons. It has a half life of 5.3 years and decays by negative beta emission to metastable 60Ni. This rapidly releases a gamma ray of either 1.17 or 1.33 MeV to reach a stable state.The source typically has an activity of 185 – 370 MBq, giving a dose rate at 80 cm of 1 – 2 Gy/minute. The Cobalt 60 source is usually replaced before a single half life has elapsed.

c)The source is rotated or slid into the treatment position when dose delivery is desired. An emergency shut off is required in the event of power failure. Dose delivery is measured by treatment time as opposed to monitor units, as the time required to deliver a dose increases as the source activity decreases. A primary timer and backup secondary timer are used to control treatment time.
Fields are defined by secondary collimators and range from 5 x 5 cm to 30 x 30 cm at an SSD of 80 cm. The geometric penumbra of the beam is dependent on the source size (limited for cobalt 60 units to about 1 cm) and is typically wider than that for a linear accelerator. penumbra trimmers, a beam accessory, can be placed closer to the patient surface to provide additional collimation and reduce the effect of the finite source size.

d)A gantry (in isocentric machines) to allow the source to rotate around a fixed position. The SAD is usually 80 or 100 cm. The gantry is typically attached to a stand which houses motors and monitoring equipment.A patient support assembly or patient couch which allows the patient to be positioned in the desired position.A bunker to protect staff, patients and the general public from unnecessary radiation exposure.A machine console outside the bunker which allows therapists to operate the machine remotely.It is important to evaluate each aspect of the equipment choice, first of all the anticipated clinical benefits to patients but also the management of the technology over time to ensure that it will deliver consistently safe and effective treatment,” said Ahmed Meghzifene, Head of the Dosimetry and Medical Radiation Physics Section at the IAEA. “If not correctly handled and maintained, radiotherapy equipment may deliver non-optimal treatments to patients and in extreme situations may cause harm.The side event ‘How to maximize the efficiency of radiation treatment – integrating cobalt-60 machines and linear accelerators’, provided a balanced perspective on these two technologies to help Member States make informed decisions based on their specific needs and conditions. Linear accelerators (linacs) and cobalt-60 (Co-60) machines are two of the most commonly used machines for external beam radiotherapy, a procedure in which high-energy beams are used to kill tumour cells.overall budget is little bit high but its the question of your health. overall it is a good choice.


Related Solutions

a tumour located at a depth of 10 cm is treated using a non-isocentic technique with...
a tumour located at a depth of 10 cm is treated using a non-isocentic technique with a Co-60 machine, to a dose of 1.8 Gy, with a single 5cm x 5cm beam. Compute the maximum dose received anywhere in the patients anatomy. b)At what depth does the maximum occur? Explain c) compute the dose at a depth of 5 cm along the beam axis d) Would this be a good strategy? Explain percent dose (in %) table for Co-60 beam...
A tumour located at a depth of 10 cm is treated using a non-isocentic technique with...
A tumour located at a depth of 10 cm is treated using a non-isocentic technique with a Co-60 machine, to a dose of 1.8 Gy, with a single 5 cm x 5 cm beam. Compute the maximum dose received anywhere in the patient's anatomy. Refer to the attached table for PDD data. a) At what depth does the maximum dose occur? b) Compute the dose at a depth of 5 cm along the beam axis
a) A tumour located at a depth of 10 cm is treated using a non-isocentic technique...
a) A tumour located at a depth of 10 cm is treated using a non-isocentic technique with a Co-60 machine, to a dose of 1.8 Gy, with a single 5 cm x 5 cm beam. Compute the maximum dose received anywhere in the patient's anatomy. Refer to the attached table for PDD data. (5 pts) b) At what depth does the maximum dose occur? Explain. (5 pts) c) Compute the dose at a depth of 5 cm along the beam...
A tumour located at a depth of 10cm is treated using a non-isocentric technique with a...
A tumour located at a depth of 10cm is treated using a non-isocentric technique with a Co-60 machine, to a dose of 1.8 Gy, with a single 5cmx5cm beam. a) Compute the maximum dose received anywhere in the patients anatomy. Refer to the attached table for PDD data. B) at what depth does the maximum dose occur? Explain c) computer the dose at a depth of 5 cm along the beam axis. d)would this be a good treatment strategy NEED...
The rectangular block below has front-face dimensions of 10 cm by 4 cm, with a depth...
The rectangular block below has front-face dimensions of 10 cm by 4 cm, with a depth of 3 cm.(Figure 1) You will be asked to treat this object as an electrical resistor. Rank the block based on its electrical resistance along the three illustrated coordinate directions (x, y, and z). Rank from largest to smallest. To rank items as equivalent, overlap them. PART B: If all of the dimensions of the block double (to become 20 cm wide, 8 cm...
In general, the index of refraction may vary with depth; this can be treated by considering...
In general, the index of refraction may vary with depth; this can be treated by considering the medium as a sequence of thin layers. a) Starting from Snell’s law obtain a relation between dθ/dz and dn/dz . [Hint: Choose a thin layer going from z to z dz + and write n andθ in terms of z and z dz + , then apply Snell’s law to either side of the layer. ] b) Apply this result to an electromagnetic...
What is depression known for? Where is located? How is it treated?
What is depression known for? Where is located? How is it treated?
A diverging lens (f = –11.0 cm) is located 24.0 cm to the left of a...
A diverging lens (f = –11.0 cm) is located 24.0 cm to the left of a converging lens (f = 35.0 cm). A 3.00-cm-tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image?
A converging lens (f1 = 24.0 cm) is located 56.0 cm to the left of a...
A converging lens (f1 = 24.0 cm) is located 56.0 cm to the left of a diverging lens (f2 = -28.0 cm). An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies 20.4 cm to the left of the diverging lens. How far is the object from the converging lens? do1 =
Clinical experience of 1 example of Therapeutic Communication Technique use and 1 Non Therapeutic Technique use (verbal or non verbal)
Clinical experience of 1 example of Therapeutic Communication Technique use and 1 Non Therapeutic Technique use (verbal or non verbal), and identify the technique that was been used . 
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT