What is the recoil velocity of an electron (in m/s), if a 44 keV photon strikes the electron initially at rest and the photon is scattered through an angle of 86°? Ответ:
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- You set up a photoelectric experiment with an unknown metal to eject electrons. You use light of wavelength λ = 670 nm, which just BARELY ejects electrons from the metal. Planck□s constant is either h = 6.63 x 10-34 J-s or h = 4.14 x 10-15 ev.s. a) What is the binding energy of the unknown metal in eV? ev b) You change the light source to one with a wavelength of λ = 310 nm. Using the binding energy you found in the previous step, find the maximum kinetic energy of an electron that is ejected from the metal in Joules. J c) What is the stopping voltage for an electron with the kinetic energy you just found? VX-rays with an energy of 400 keV undergo Compton scattering from a target. The scattered rays are deflected at an angle of 32.0° relative to the direction of the incident rays. What is the energy of the scattered x-rays? a. 3.58E+2 keV b. 4.00E+2 keV c. 3.37E+3 keV d. 3.23E+2 keVFor light with a wavelength of 350 nm and with an intensity of /= 10-8 W/m², what is the number of photons/(m²s) in the light beam?
- If a 7.0-keV photon scatters from a free proton at rest, what is the change in the photon's wavelength if the photon recoils at 90°?A photon of wavelength 0.90638 nm strikes a free electron that is initially at rest. The photon is scattered straight backward. What is the speed of the recoil electron after the collision? V = i ! m/sA x-ray having wavelength 0.077nm strikes a free and ‘stationary’ electron. This causes the electron to recoil and an x-ray having longer wavelength to be emitted at angle θ to the direction of the incident x-ray. If the electron recoils with kinetic energy 0.9238 eV, determine the angle at which the second x-ray emerges