X-rays with a wavelength of 1.0 nm are scattered from two small tumors in the human body.If the two tumors are a distance of 5.0 cm away from the x-ray detector,and the detector has an entrance aperture of 0.5 mm,what is the minimum separation between the two tumors that will allow the x-ray detector to determine that there are two tumors instead of one?

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Quizplus · Accepted Answer

The minimum separation between the two tumors that will allow the x-ray detector to determine that there are two tumors instead of one is given by the Rayleigh criterion:$$\Delta x = 1.22 \lambda \frac{d}{a}$$where:$$\Delta x$$ is the minimum separation between the two tumors$$\lambda$$ is the wavelength of the x-rays$$d$$ is the distance between the tumors and the detector$$a$$ is the diameter of the entrance aperture of the detectorPlugging in the given values, we get:$$\Delta x = 1.22 (1.0 	imes 10^{-9} 	ext{ m}) \frac{0.05 	ext{ m}}{0.0005 	ext{ m}} = 1.22 	imes 10^{-7} 	ext{ m} = 122 	ext{ nm}$$Therefore, the minimum separation between the two tumors that will allow the x-ray detector to determine that there are two tumors instead of one is 120 nm.

X-Rays with a Wavelength of 1