Relativistic de Broglie wavelength

Question

Suppose a particle of mass m in special relativity. Suppose the de Broglie wave-length is (non-relativistic) case:
$$lambda=dfrac{h}{p}=dfrac{h}{mv}$$
In the case of RELATIVISTIC particle, the momentum is $p=mgamma v$. Therefore a way to recast the de Broglie wavelength is:
$$lambda_{r}=dfrac{hsqrt{1-v^2/c^2}}{mv}$$
Suppose now that we focus on the kinetic energy. For a free particle, we get in the nonrelativistic case, $K.E.=T=p^2/2m$, and thus $p=sqrt{2Tm}$, and so
$$lambda=dfrac{h}{sqrt{2Tm}}$$
I have a doubt concerting to the relativistic case. The natural election for the de Broglie wave-length in the relativistic case is well known: you take $T=E-mc^2$, and from $E^2=(pc)^2+(mc^2)^2$, by simple substitution of $E=T+mc^2$, you get $(pc)^2=T^2+2Tmc^2$, $$p=sqrt{T^2/c^2+2Tm}=sqrt{2Tm(1+T/2mc^2)}$$
or 
$$lambda=dfrac{h}{sqrt{T^2/c^2+2Tm}}=dfrac{h}{sqrt{2Tm(1+T/2mc^2)}}$$
Well, I have seen a couple of places where the relativistic de Broglie wavelength for a kinetic colliding partice is assumed to be
$$lambda=dfrac{hc}{T}=dfrac{hc}{sqrt{(pc)^2+(mc^2)^2}-mc^2}$$
Is this last relativistic consistent in certain limit (it seems is the ultra-relativistic case) to the previous one?

riemannium · Answer

Thinking about this answer:
$$KE(rel)=(gamma-1)mc^2$$
Thus, if T=KE(rel):
$$gamma=1+T/mc^2$$
If $T>>2mc^2>mc^2$:
$$gammaapprox T/mc^2$$
By the other hand, from the de Broglie fundamental relationship:
$$lambda=dfrac{h}{p}=dfrac{hc}{pc}=dfrac{hc}{mgamma v c}$$
and then, with $gammaapprox T/mc^2$, and $vapprox c$ we obtain
$$lambda=dfrac{h mc^3}{m T v c}=dfrac{h c}{T}$$
From this last equation we get the above one from a simple use of the dispersion relationship. However, as the kinetic energy is much bigger than the rest mass, my doubt concerning the presence of mass remained...Until I did some numbers...

Relativistic de Broglie wavelength

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