[Physics] 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 $p=m\gamma v$. Therefore a way to recast the de Broglie wavelength is:
$$\lambda_{r}=\dfrac{h\sqrt{1-v^2/c^2}}{mv}$$
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?

Answer 1

Thinking about this answer: $$KE(rel)=(\gamma-1)mc^2$$ Thus, if T=KE(rel): $$\gamma=1+T/mc^2$$ If $T>>2mc^2>mc^2$: $$\gamma\approx T/mc^2$$ By the other hand, from the de Broglie fundamental relationship: $$\lambda=\dfrac{h}{p}=\dfrac{hc}{pc}=\dfrac{hc}{m\gamma v c}$$ and then, with $\gamma\approx T/mc^2$, and $v\approx c$ 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...