The continuous spectrum from an X-ray tube has a short-wavelength cutoff at 0.031 nm. Take hc = 1240 eV·nm. If the accelerating potential is approximately N kV, find N.
Physics · Atomic Structure and X-rays · x-rays and its types
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Question

The continuous spectrum from an X-ray tube has a short-wavelength cutoff at 0.031 nm. Take hc = 1240 eV·nm. If the accelerating potential is approximately N kV, find N.

G1 Physics · Accepted Answer

At the cutoff wavelength, eV = hc ÷ λ(min). Hence the electron energy in electron-volts is numerically equal to the accelerating voltage in volts. E(max) = 1240 ÷ 0.031 eV = 40,000 eV = 40 keV. Therefore the required accelerating potential is 40 kV. Hence N = 40.