Dissertativa 10 - 2ª Fase - Dia 3 - ITA 2027

a) Sejam ${\mathrm{E}}_{\mathrm{e}}$ e ${\mathrm{p}}_{\mathrm{e}}$ a energia e o módulo do momento do elétron após a colisão. Conservando a energia:

$\mathrm{h\nu }+{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2=\mathrm{h\nu }\text{'}+{\mathrm{E}}_{\mathrm{e}}$

${\mathrm{E}}_{\mathrm{e}}=\mathrm{h}(\mathrm{\nu }-\mathrm{\nu }\text{'})+{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2$

Pela conservação do momento, os momentos formam o triângulo da figura. Pela lei dos cossenos:

${\mathrm{p}}_{\mathrm{e}}^2={\left(\frac{\mathrm{h\nu }}{\mathrm{c}}\right)}^2+{\left(\frac{\mathrm{h\nu }\text{'}}{\mathrm{c}}\right)}^2-2\left(\frac{\mathrm{h\nu }}{\mathrm{c}}\right)\left(\frac{\mathrm{h\nu }\text{'}}{\mathrm{c}}\right)\mathrm{cos\theta }$

Da dinâmica relativística, vem a relação entre a energia e o momento do elétron:

${\mathrm{E}}_{\mathrm{e}}^2={\mathrm{p}}_{\mathrm{e}}^2{\mathrm{c}}^2+{\mathrm{m}}_{\mathrm{e}}^2{\mathrm{c}}^4$

Assim:

${\mathrm{p}}_{\mathrm{e}}^2{\mathrm{c}}^2=\left[\mathrm{h}\right(\mathrm{\nu }-\mathrm{\nu }\text{'})+{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2{]}^2-{\mathrm{m}}_{\mathrm{e}}^2{\mathrm{c}}^4$

Na equação da colisão:

$\left[\mathrm{h}\right(\mathrm{\nu }-\mathrm{\nu }\text{'})+{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2{]}^2-{\mathrm{m}}_{\mathrm{e}}^2{\mathrm{c}}^4={\mathrm{h}}^2{\mathrm{\nu }}^2+{\mathrm{h}}^2\mathrm{\nu }{\text{'}}^2-2{\mathrm{h}}^2\mathrm{\nu \nu }\text{'}\mathrm{cos\theta }$

${\mathrm{h}}^2(\mathrm{\nu }-\mathrm{\nu }\text{'}{)}^2+2\mathrm{h}(\mathrm{\nu }-\mathrm{\nu }\text{'}){\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2={\mathrm{h}}^2{\mathrm{\nu }}^2+{\mathrm{h}}^2\mathrm{\nu }{\text{'}}^2-2{\mathrm{h}}^2\mathrm{\nu \nu }\text{'}\mathrm{cos\theta }$

$-2{\mathrm{h}}^2\mathrm{\nu \nu }\text{'}+2\mathrm{h}(\mathrm{\nu }-\mathrm{\nu }\text{'}){\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2=-2{\mathrm{h}}^2\mathrm{\nu \nu }\text{'}\mathrm{cos\theta }$

${\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2(\mathrm{\nu }-\mathrm{\nu }\text{'})=\mathrm{h\nu \nu }\text{'}(1-\mathrm{cos\theta })$

Isolando $\mathrm{\nu }\text{'}$$\Rightarrow \boxed{\nu \text{'}=\frac{\nu }{1+\frac{h\nu }{m_e{c}^2}(1-\cos \theta )}}$

b) Como $\mathrm{\nu }\text{'}<\mathrm{\nu }$, o observador deve se mover na direção do fóton espalhado, em sentido oposto ao de sua propagação, para medir uma frequência maior. Tomando $\mathrm{\beta }=\frac{\mathrm{V}}{\mathrm{c}}$

$\mathrm{\nu }=\mathrm{\nu }\text{'}\sqrt{\frac{1+\mathrm{\beta }}{1-\mathrm{\beta }}}\Rightarrow \mathrm{\beta }=\frac{{\mathrm{\nu }}^2-\mathrm{\nu }{\text{'}}^2}{{\mathrm{\nu }}^2+\mathrm{\nu }{\text{'}}^2}$

Do item a:

$\mathrm{\beta }=\frac{{\mathrm{\nu }}^2-{\left(\frac{\mathrm{\nu }}{1+\frac{\mathrm{h\nu }}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}(1-\mathrm{cos\theta })}\right)}^2}{{\mathrm{\nu }}^2+{\left(\frac{\mathrm{\nu }}{1+\frac{\mathrm{h\nu }}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}(1-\mathrm{cos\theta })}\right)}^2}$

$\mathrm{\beta }=\frac{1-\frac{1}{{\left(1+\frac{\mathrm{h\nu }}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}(1-\mathrm{cos\theta })\right)}^2}}{1+\frac{1}{{\left(1+\frac{\mathrm{h\nu }}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}(1-\mathrm{cos\theta })\right)}^2}}$

$\mathrm{\beta }=\frac{{\left(1+\frac{\mathrm{h\nu }}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}(1-\mathrm{cos\theta })\right)}^2-1}{{\left(1+\frac{\mathrm{h\nu }}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}(1-\mathrm{cos\theta })\right)}^2+1}$

Finalmente, com $\mathrm{V}=\mathrm{\beta c}$$\Rightarrow \boxed{\mathrm{V}=\left[\frac{{\left(1+\frac{\mathrm{h\nu }}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}(1-\mathrm{cos\theta })\right)}^2-1}{{\left(1+\frac{\mathrm{h\nu }}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}(1-\mathrm{cos\theta })\right)}^2+1}\right]\mathrm{c}}$