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Chapter 6: Problem 157

Find the de Broglie wavelength of an electron accelerated from rest in an X-ray tube in the potential difference of \(100 \mathrm{keV}\). (Rest mass energy of an electron is \(E_{0}=511 \mathrm{keV}\).

Short Answer

Expert verified
The de Broglie wavelength of an electron accelerated from rest in an X-ray tube with a potential difference of 100 keV can be calculated using the following steps: 1. Convert the potential difference to energy (1.6 × 10^{-14} J) 2. Calculate relativistic energy 3. Determine relativistic mass 4. Find the momentum of the electron using the energy-momentum relation 5. Calculate the de Broglie wavelength using the formula \(\lambda = \frac{h}{p}\), where h is Planck's constant After performing these calculations, we can find the de Broglie wavelength of the electron in meters.

Step by step solution

01

Converting potential difference to energy

First, we need to convert the potential difference of 100 keV to energy, which can be done using the following formula: Energy \(= qV\) where \(q\) is the charge of the electron \((-1.6 \times 10^{-19} C)\) and \(V\) is the potential difference (100 keV = \(10^{5} eV\)). Energy = \((-1.6 \times 10^{-19})(10^{5})\) Energy = \(1.6 \times 10^{-14} J\) The kinetic energy of the electron is 1.6 \( \times 10^{-14} J\).
02

Calculate relativistic energy

Since the kinetic energy of the electron is significant compared to its rest mass energy (511 keV), we need to use the relativistic energy expression: Relativistic energy = \((1.6 \times 10^{-14} J) + (511 \times 10^3 eV)(1.6 \times 10^{-19} J/eV) = E\)
03

Determine relativistic mass

Next, we can use the energy-momentum relation to find the relativistic mass of the electron: \(E = mc^2\) Therefore, the relativistic mass \(m = \frac{E}{c^2}\) #c^2 = 9 × 10^{16} m^2/s^2#
04

Find the momentum of the electron

Now that we have the relativistic mass, we can compute the momentum using the following expression: \( p = mv \) To relate the momentum with energy, we can use the following equation: \(E^2 = (mc^2)^2 + (pc)^2\) Rearranging the equation for momentum: \(p = \frac{1}{c} \sqrt{E^2 - (mc^2)^2}\)
05

Calculate the de Broglie wavelength

Finally, using the de Broglie wavelength formula: \(\lambda = \frac{h}{p}\) where h is Planck's constant = \(6.63 \times 10^{-34}Js\) \(\lambda = \frac{6.63 \times 10^{-34}}{p}\) After substituting the value of momentum from step 4 and solving for the de Broglie wavelength, we will find the wavelength of the electron in the X-ray tube in meters.

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