Question

If the shortest-wavelength bremsstrahlung X-rays emitted from an X-ray tube have λ=0.030nm,\lambda=0.030 \mathrm{nm}, what is the voltage across the tube?

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The energy of an electron in the tube is given by:

E=hcλ=6.626×1034 m2 kg/s3×108 m/s3×1011 m=6.626×1015 JE = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \text{ m$^2$ kg/s} \cdot 3 \times 10^8 \text{ m/s}}{3 \times 10^{-11} \text{ m}} = 6.626 \times 10^{-15} \text{ J}

This means that the voltage in the tube is:

U=Ee=6.626×1015 J1.6×1019 C=41.4 kVU = \frac{E}{e} = \frac{6.626 \times 10^{-15} \text{ J}}{1.6 \times 10^{-19} \text{ C}} = \boxed{41.4 \text{ kV}}

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