Exam 24: Photon Dosimetry Concepts and Calculations

A patient is being treated on the 18-MV beam to a PA spine at a depth of 10 cm.The collimator setting is 10 x 13 with no blocks.The dose is 300 cGy per fraction.What is the dose delivered to Dmax?

A TAR at the depth of Dmax used to correct for scatter of dose is called the _________.

A wedge filter is used to alter the isodose distribution on an anterior chest field.The MU setting must be ________________ to account for the wedge filter in the field.

Adjacent fields that have an overlap at depth due to divergence are called:

A patient is treated on a 6-MV machine at 100 cm SSD.The collimator setting is 15 x 15.There is no blocking used for this treatment.The prescription states that a dose of 3000 cGy is to be delivered to a depth of 3 cm in 10 fractions using a posterior treatment field using the nonisocentric method of calculation.The patient central axis separation is 20 cm.What is the MU setting for the treatment? Reference dose rate (RDR)for a 15 cm equivalent square is 0.993 cGy/MU.

The equivalent square of rectangular field of 10 x 15 is:

As the field size increases,the TAR:

All of the following apply to "absorbed dose" except:

Find the PDD at 10 cm of a 6-MV beam of radiation that has a 5 x 5 cm treatment field.

As the field size increases,the PDD:

A patient is being treated on the 6-MV beam to an opposed lateral whole brain (midline).The collimator setting is 14 x 18 with no blocks.The dose is 4000 cGy/20 fractions.The separation is 22 cm.What is the dose delivered to Dmax?

Absorbed dose at depth x 100% = absorbed dose at Dmax is the definition of:

What is the patient separation if depth from the AP is 9 cm,and depth from PA is 11 cm?

According to the text,methods of obtaining dose uniformity across field junctions include: I.dosimetric isodose matching II.separated fields III.junction shift IV.half beam blocking V.geometric matching

A patient is being treated on the 18-MV beam to an AP/PA stomach (midline).The collimator setting is 12 x 12 with no blocks.The dose is 4400 cGy/22 fractions.The separation is 24 cm.The cord is 4 cm from the PA.What is the dose delivered to the spinal cord?

As the SSD increases,the PDD:

Dmax is defined as:

Find the skin gap necessary to match two symmetric ports at midline that are 21 and 37 cm in length.Assume the midline depth is 13 cm and both ports are treated at 100 cm SAD.

As a field size increases from the standard established,the output factor for a treatment machine will _______________.

Find the PDD at 3.5 cm of an 18-MV beam of radiation that has a 23 x 15 cm treatment field.