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A 0.001 60-nm photon scatters from a free electron.

For what (photon) scattering angle does the recoiling electron have kinetic energy equal to the energy of the scattered

photon?

70.0^{\circ}

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Cornell University

University of Michigan - Ann Arbor

Hope College

University of Winnipeg

The solution for the answer is that it has given dead the wavelength off the incident photon, that is Lambda Note is equal to 0.16 and m the energy off the incident photon is equally divided into the boiling electron and the scared on if the kinetic energy off the recalling electrons is equal to the energy off the scattered four tone. So the energy off the scattered photon is equal to the energy of incident. Photon divided by two now so E s R is equal toe e i n gamma by two at Jeff equal toe f no divided by two. So this will be equal toe Lambda is equal to two lambda Note The shift in the wavelength off the scattered photon is given by as Delta Lambda called to Lambda minus lambda Note, which is equal to act by M. E. C. One minus cost hidta and this will be equal to lambda Note equal toe h m E C. One minus cost heater. And by simplifying this, we get cost it a call to one minus M e c. Lambda Nordby edge Here it is, given that m e is the mosque often. Elect Undated 9.11 10 to the power minus 31 kg, C is equal to the speed of light that is three multiply by 10 to the power 8 m per second at is the plank's constant. It is six 0.63 multiplied by identity, power minus or deport Js Here we substitute the values in the above equation and this and by solving this we get 0.341 So t tickle toe cause inverse 0.341 and it is 70 degrees. And this is our answer. So this is the solution. Step by step in your tail. Please go through this. Thank you.

DR APJ ABDUL KALAM TECHNICAL UNIVERSITY