After a 0.800-nm x-ray photon scatters from a free electron, the electron recoils at 1.40 × 10 6 m / s. What is the scattering angle?

Question

After a 0.800-nm x-ray photon scatters from a free electron, the electron recoils at   1.40  ×      10    6       m      /    s. What is the scattering angle?

Davian Nguyen · Accepted Answer

We know Compton shift is given by,  △  λ  =      h                  m        e            c        (  1  −  cos  ⁡  θ  )  Solving for   θ we get,  (  1  −  cos  ⁡  θ  )  =                    m        e            c      △      λ        h      θ  =      cos          −      1        ⁡  (  1  −                    m        e            c      △      λ        h    )Substitute   △  λ  as 0.003 nm, h as   6.626  ×      10          −      34           J  ⋅  s as   9.1  ×      10          −      31           k  g, and  c as   3  ×      10    8       m  ⋅      s          −      1        θ  =      cos          −      1        ⁡  (  1  −            (      9.1      ×              10                  −          31                           k      g      )      (      3      ×              10        8            m      ⋅              s                  −          1                    )      (      0.003      n      m      )              (      6.626      ×              10                  −          34                           J      ⋅      s      )        )    =      cos          −      1        ⁡  (  −  0.236  )    =      103.3    ∘  Thus the angle of scattering will be       103.3    ∘  .

After a 0.800-nm x-ray photon scatters from a free electron, the electron recoils at 1.40 xx 10^6 m/s. (b) What is the scattering angle?

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