Question

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

Solution

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

$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \text{ m^2 kg/s} \cdot 3 \times 10^8 \text{ m/s}}{3 \times 10^{-11} \text{ m}} = 6.626 \times 10^{-15} \text{ J}$

This means that the voltage in the tube is:

$U = \frac{E}{e} = \frac{6.626 \times 10^{-15} \text{ J}}{1.6 \times 10^{-19} \text{ C}} = \boxed{41.4 \text{ kV}}$

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