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Chapter 27: Problem 17

What is the cutoff frequency for an x-ray tube operating at \(46 \mathrm{kV} ?\)

Short Answer

Expert verified
Answer: The cutoff frequency for the x-ray tube operating at 46 kV is 1.11 x 10^18 Hz.

Step by step solution

01

Convert voltage to energy

To convert the given voltage (46 kV) to energy, we first convert it to volts and then use the formula E = qV, where E is the energy, q is the electric charge of an electron, and V is the voltage. The charge of an electron is approximately \(1.6 x 10^{-19} \mathrm{C}\). Voltage in volts: \(46\times10^3 \mathrm{V}\) Energy: \(E = qV = (1.6\times10^{-19}\,\mathrm{C})(46\times10^3\,\mathrm{V}) = 7.36\times10^{-16}\, \mathrm{J}\)
02

Use the energy-frequency relationship

The energy-frequency relationship is given by the equation \(E = h\nu\), where E is the energy, h is the Planck's constant, and \(\nu\) is the frequency of the x-ray photon. Planck's constant is approximately \(6.63\times10^{-34}\,\mathrm{Js}\). We can solve for the frequency as follows: \(\nu = \frac{E}{h} = \frac{7.36\times10^{-16}\,\mathrm{J}}{6.63\times10^{-34}\,\mathrm{Js}}\)
03

Calculate the cutoff frequency

Now we can plug in the values to calculate the cutoff frequency, \(\nu\): \(\nu = \frac{7.36\times10^{-16}\,\mathrm{J}}{6.63\times10^{-34}\,\mathrm{Js}} = 1.11\times10^{18}\, \mathrm{Hz}\) The cutoff frequency for the x-ray tube operating at \(46\,\mathrm{kV}\) is \(1.11\times10^{18}\, \mathrm{Hz}\).

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