A hydrogen atom is in its ground state when its electronabsorbs 48 eV in an inte

Question

A hydrogen atom is in its ground state when its electronabsorbs 48 eV in an interaction with a photon. What is theenergy of the resulting free electron? Additionally, suppose also that the free electron moves alongthe original line of the incoming photon. Obviously thesolution to the problem shows that energy is conserved in thiscollision between an electron tnd photon. Is momentumconserved? If not, where does the departure betweeen originaland final momentum probably go? A hydrogen atom is in its ground state when its electronabsorbs 48 eV in an interaction with a photon. What is theenergy of the resulting free electron? Additionally, suppose also that the free electron moves alongthe original line of the incoming photon. Obviously thesolution to the problem shows that energy is conserved in thiscollision between an electron tnd photon. Is momentumconserved? If not, where does the departure betweeen originaland final momentum probably go?

Explanation / Answer

the energy of the resulting free electron E = E1 + Eo E1 = -13.6 eV and Eo= 48 eV or E = -13.6 + 48 = 34.4 eV the x and y components of linear momentum are x component: (h/o) = (h/') * cos +meu * cos y component: 0 = (h/') * sin - meu *sin Here, = = 0o x component: x component: or (h/o) = (h/') * cos(0o) +meu * cos(0o) or (h/o) = (h/') + meu----------------(1) y component: 0 = (h/') * sin(0o) - meu *sin(0o) = 0 -----------------(2) where = (1/(1 - u^2/c^2)^(1/2)) and u is the speed ofthe electron therefore,the momentum is not conserved from equation(1) (h/o) = (h/') +meu or (h/o) - (h/') =meu the departure betweeen original and final momentum goes intothe momentum of the recoiling electron.

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