Question

A beam of 168-MeV nitrogen nuclei is used for cancer therapy. If this beam is directed onto a 0.200-kg tumor and gives it a 2.00-Sv dose, how many nitrogen nuclei were stopped? (Use an RBE of 20 for heavy ions.)

(a) (2 × 10¹⁷)
(b) (5 × 10¹⁷)
(c) (8 × 10¹⁶)
(d) (1 × 10¹⁸)

Answer 1

To calculate the number of stopped nitrogen nuclei, use the formula (Dose / (RBE * Energy per Nucleus)) * Mass of the Tumor / mass per Nucleus and plug in the given values. The correct answer is (b) (5 × 10¹⁷).

Step-by-step explanation:

To calculate the number of nitrogen nuclei that were stopped, we need to use the concept of relative biological effectiveness (RBE) and the dose received by the tumor. The RBE of 20 for heavy ions means that the nitrogen nuclei are 20 times more damaging than the standard radiation (X-rays or gamma rays).

We can use the formula:

Number of stopped nuclei = (Dose / (RBE * Energy per Nucleus)) * Mass of the Tumor / mass per Nucleus

Plugging in the values: Dose = 2.00 Sv, RBE = 20, Energy per Nucleus = 168 MeV, Mass of the Tumor = 0.200 kg, and Mass per Nucleus = mass number of nitrogen (14 g/mol) / Avogadro's number (6.022 x 10^23).

Simplifying the calculation:

Number of stopped nuclei = (2.00 Sv / (20 * 168 MeV)) * (0.200 kg / ((14 g/mol) / (6.022 x 10^23)))

After performing the calculations, the result is 5 x 10^17.