A 30 keV electron is brought to rest by colliding twice with target nuclei as in

Question

A 30 keV electron is brought to rest by colliding twice with target nuclei as in the figure below. (Assume the nuclei remain stationary.) The wavelength associated with the photon emitted in the second collision is 110 pm greater than that associated with the photon emitted in the first collision. What is the kinetic energy of the electron after the first collision? X keV What is the wavelength and energy of the first photon? lambda 1 = pm E1 = keV What is the wavelength and energy of the second photon? lambda 2 = pm E2 = keV

Explanation / Answer

KE after 1st collision=hc/(lemda+110)=7.4keV hc/lemda+hc/(lemda+110)=30*1.6*10^-16 lemda1=54.76pm E1=hc/lemda =22.6keV

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