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14.07.2022 - 23:29

A beam of 35.0 keV electrons strikes a molybdenum target, generating the X -rays. What is the cutoff wavelength?

Question:

A beam of {eq}35.0 keV {/eq} electrons strikes a molybdenum target, generating the {eq}X {/eq} -rays. What is the cutoff wavelength?

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Eddie April 3, 2023 в 21:17

The cutoff wavelength can be calculated using the equation: ? = hc/E Where ? is the cutoff wavelength, h is Planck's constant, c is the speed of light, and E is the energy of the electrons in electronvolts (eV). First, we need to convert the energy of the electrons from keV to eV: 35.0 keV = 35,000 eV Then, we can plug in the values and solve for ?: ? = hc/E ? = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(35,000 eV x 1.60 x 10^-19 J/eV) ? = 0.071 nm Therefore, the cutoff wavelength of the X-rays generated by the beam of 35.0 keV electrons striking a molybdenum target is 0.071 nm.

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