A material whose K absorption edge is 0.15 Å is irradiated with 0.1 Å X-rays. The maximum kinetic energy of the photoelectrons that are emitted from the K-shell is :
Correct answer: A. 41 keV
- A. 41 keV
- B. 51 keV
- C. 61 keV
- D. 71 keV
Explanation
To determine the maximum kinetic energy of the photoelectrons, we need to calculate the energy of the incident X-rays and the energy corresponding to the K absorption edge. The energy of a photon is given by E = hc/λ, where h is Planck's constant and c is the speed of light.The energy of the 0.1 Å X-rays is calculated as:Eincident = (6.626 x 10-34 J·s)(3 x 108 m/s) / (0.1 x 10-10 m) = 1.986 x 10-15 J = 124 keVThe energy corresponding to the K absorption edge is:EK = (6.626 x 10-34 J·s)(3 x 108 m/s) / (0.15 x 10-10 m) = 1.324 x 10-15 J = 83 keVThe maximum kinetic energy of the photoelectrons is the difference:Ekinetic = Eincident - EK = 124 keV - 83 keV = 41 keVThus, the correct answer is 41 keV. The other options represent incorrect calculations of this energy difference, leading to overestimated or inconsistent values.
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Classical physics cannot explain blackbody radiation and related observations, leading to Planck’s quantum theory, in which energy is emitted or absorbed in discrete packets. Photons provide the particle model of light, with energy and momentum linked to frequency and wavelength, including the photoelectric effect and its threshold frequency.
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