Question

In a X-Ray tube operating at 20 kV, the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of the generated X-rays is ________ (rounded off to two decimal places).
Given: e/m ratio for an electron = 1.76×10¹¹ Ckg-1 and the speed of light in vacuum is 3×10⁸ ms ^-1

Answer 1

Correct Answer: 0.12 - 0.16

To solve the problem, we first need to understand the two wavelengths involved: the de-Broglie wavelength of incident electrons and the shortest wavelength of generated X-rays. 
The de-Broglie wavelength (\( \lambda_{\text{electron}} \)) for an electron is given by:
\[ \lambda_{\text{electron}} = \frac{h}{mv} \] where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)), \( m \) is the mass of the electron, and \( v \) is the velocity of the electron.
Using the relation \( eV = \frac{1}{2}mv^2 \) (kinetic energy), where \( V \) is the accelerating voltage (\( 20 \, \text{kV} = 20,000 \, \text{V} \)), we have: \[ v = \sqrt{\frac{2eV}{m}} \] The de-Broglie wavelength becomes: \[ \lambda_{\text{electron}} = \frac{h}{\sqrt{2meV}} \] Now, for the shortest wavelength of X-rays (\( \lambda_{\text{min}} \)), we use the equation from X-ray generation: \[ \lambda_{\text{min}} = \frac{hc}{eV} \] where \( c \) is the speed of light (\( 3 \times 10^8 \, \text{ms}^{-1} \)), and \( h \) and \( e \) are constants.
Thus: \[ \lambda_{\text{min}} = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{ms}^{-1})}{(1.6 \times 10^{-19} \, \text{C}) \times (20,000 \, \text{V})} \] \[ \lambda_{\text{min}} \approx 6.21 \times 10^{-11} \, \text{m} \] Next, we calculate the de-Broglie wavelength using: \[ \lambda_{\text{electron}} = \frac{6.626 \times 10^{-34} \, \text{Js}}{\sqrt{2 \times 9.11 \times 10^{-31} \, \text{kg} \times 1.6 \times 10^{-19} \, \text{C} \times 20,000 \, \text{V}}} \] Using \( \frac{e}{m} = 1.76 \times 10^{11} \, \text{C/kg} \), we find: \[ \lambda_{\text{electron}} \approx 8.67 \times 10^{-12} \, \text{m} \] Finally, the ratio \( \frac{\lambda_{\text{electron}}}{\lambda_{\text{min}}} \) is: \[ \frac{8.67 \times 10^{-12} \, \text{m}}{6.21 \times 10^{-11} \, \text{m}} \approx 0.14 \] Thus, the ratio of the de-Broglie wavelength to the shortest X-ray wavelength is 0.14, which falls within the provided range [0.12, 0.16].