# Which of the following two formulas is the correct wavelength expression of a relativistic electron? [closed]

$$\lambda=\dfrac{h}{m\gamma v } \tag{01}\label{01}$$ or $$\lambda=\dfrac{hv}{m\gamma c^2 } \tag{02}\label{02}$$

Edit: The energy of an electron is, by special relativity $$E=\gamma mc^2 \tag{03}\label{03}$$ where $$\gamma = \left(1-\frac{v^2}{c^2}\right)^{-1/2} \tag{04}\label{04}$$ The frequency $$f$$ satisfies the Planck relation: $$E = hf \tag{05}\label{05}$$ and the frequency and wavelength $$\:\lambda\:$$ of any wave must satisfy $$f\lambda = v \tag{06}\label{06}$$ Put all of those together and I get $$\lambda = \dfrac{v}{f} = \dfrac{hv}{E} = \dfrac{hv}{\gamma mc^2} \tag{07}\label{07}$$

• Commented Feb 6, 2021 at 20:01
• It seems to be the first one, so where am I wrong:
– user288235
Commented Feb 6, 2021 at 20:24

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The correct wavelength expression of a relativistic electron is (01), while (02) is wrong. This is due to the wrong equation (06). The wavelength $$\:\lambda\:$$ and the frequency $$\:f\:$$ are characteristics of the "superluminal" plane phase wave associated to the "subluminal" particle. The speed of this phase wave is $$\:c^2/\upsilon$$, so the correct equation (06) is $$$$f\lambda\boldsymbol{=}\dfrac{c^2}{\upsilon} \tag{A-01}\label{A-01}$$$$ Hence the correct equation (07) is $$$$\lambda \boldsymbol{=}\dfrac{c^2}{f\upsilon}\boldsymbol{=}\dfrac{c^2}{\left(E/h\right)\upsilon}\boldsymbol{=}\dfrac{c^2}{\left(\gamma m c^2/h\right)\upsilon}\boldsymbol{=}\dfrac{h}{\gamma m \upsilon} \tag{A-02}\label{A-02}$$$$ that is equation (01) of the question.