Sunday 14 November 2021

Integral dose (ID)

This is the total energy absorbed/deposited by ionizing radiation, within a body/in the treated volume (in J = kg × Gy) [1]. It is «one way of comparing dose distributions for different-quality beams. (...) If a mass of tissue receives a uniform dose, then the integral dose is simply the product of mass and dose. However, in practice, the absorbed dose in the tissue is nonuniform so rather complex mathematical formulas are required to calculate it. For a single beam of x- or γ radiation, Mayneord formulated the following expression:

Σ=1.44 D0  A d1⁄2 (1 - e^(0.693 d/d1⁄2))  (1 + (2.88 d1⁄2)/SSD)

where ∑ is the integral dose, D0 is the peak dose along the central axis, A is the geometric area of the field, d is the total thickness of the patient in the path of the beam, d1⁄2 is the half-value depth or the depth of 50% depth dose, and SSD is the source to surface distance. The term (1 + (2.88 d1⁄2)/SSD) is a correction for the geometric divergence of the beam. Because the integral dose is basically the product of mass and dose, its unit is the kilogram-gray or simply joule (since 1 Gy = 1 J/kg). (...) It is generally believed that the probability of damage to normal tissue increases with the increase in the integral dose, (...) [2].»
Bibliographic references:
[1] Beyzadeoglu, M., Ozyigit, G. and Ebruli, C., 2010. Basic radiation oncology. 1st ed. Berlin: Springer, p.23.
[2] Khan, F. and Gibbons, J., 2016. Khan's the physics of radiation therapy. 5th ed. Philadelphia: Wolters Kluwer, p.181.

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