※ 引述《cscscscs22 (QQQ)》之銘言：
: A 80Mev photon is elastically scattered by a proton that is at rest
: what's the max possible energy loss for the photon
: (the compton wavelenth of electron is 2.4 X10^-12)
: Ans.68.5Mev
: 算了老半天算不出來
: 哭哭
λ'-λ=h/Mc(1-cosθ) cosθ=-1時λ'最大，ν'最小→光子能量最小，損失最大
得λ'=λ+2h/Mc
碰撞後光子能量
右邊上下同除λ
E'=hν'=hc/λ'=(hc/λ)(λ/λ')=(hc/λ)[λ/(λ+2h/Mc)]=(hc/λ)[1/(1+(2h/Mcλ))]
通分乘上c → =(hc/λ)[1/(1+(2hc/Mc^2λ))]=(pc)[1/(1+2pc/Mc^2)] (1)
碰撞前 E=hν=pc=80Mev ，質子靜止能量 Mc^2=938Mev
代入(1)式可得E'=68.34Mev △E=E-E'=80-68.34=11.66Mev
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