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I tried to combine the mass–energy equivalence for a particle with mass,

$$E=\sqrt{(mc^2)^2+(pc)^2}=\sqrt{(mc^2)^2+(\gamma mvc)^2}$$

with de Broglie wavelength,

$$\lambda=\dfrac{h}{p}=\dfrac{h}{\gamma mv}. $$

I get this equation:

$$E=\dfrac{hc^2}{\lambda v}.$$

This does not seem right, since the equations suggest the energy increases as the speed slow down which is not the case. But I can't see what I did wrong, either. Can someone help me?

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  • $\begingroup$ When I do this substitution, I don't get a fraction. Could you show more of your working? Exactly how did you try to combine these equations? $\endgroup$ – perilousGourd Jul 28 '17 at 22:10
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Your algebra is correct: $$ \frac1{\lambda v} = \frac {\gamma m}h \\ \frac{hc^2}{\lambda v} = \gamma m c^2 = E $$ Your interpretation,

the equations suggest the energy increases as the speed slow down, which is not the case

is not correct because both $v$ and $\lambda$ change as energy and velocity increase.

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The wavelength $\lambda$ cannot stay constant if the velocity $v$ changes. The question is "do they change proportionally or not?" A quick examination of your deBroglie wavelength relationship shows the following: $$\lambda v = \frac{h}{\gamma m}.$$ We know that $\gamma$ is always greater than or equal to 1 and increases non-linearly with increasing $v$. Therefore, your denominator in the final expression of $E$ is decreasing non-linearly as $v$ increases, resulting in an increasing energy.

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