is $p=\frac{h}{\lambda}$ only true for massless particles? because generally $E=\sqrt{p^2c^2+m^2c^4}$, then if we equate it to $h\nu$ we get $$p=\sqrt{\frac{h^2}{\lambda^2}-m^2c^2}\neq\frac{h}{\lambda}$$
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1$\begingroup$ See this previous question and answers physics.stackexchange.com/q/139641 $\endgroup$– GiorgioP-DoomsdayClockIsAt-90Commented Apr 5, 2021 at 16:33
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$\begingroup$ You seem to be assuming that $\lambda\nu=c$, which is not true for massive particle d-B waves. $\endgroup$– mike stoneCommented Apr 5, 2021 at 16:38
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Relativistic energy-momentum relation: $$p_\mu p^\mu = (p^0)^2 - \vec p^2= m_0^2 c^2$$ where $$p^0 c=E=\hbar \omega \qquad p^i = \hbar k^i$$ subject to $|k|=2\pi/\lambda$ and $\omega = 2\pi\nu$.
Therefore, generally $$|p| = \sqrt{\frac{h^2 \nu^2}{c^2}-m_0^2 c^2}=\frac{h}{\lambda}$$
Only for a massless particle ($m_0=0$) the energy-momentum relation reduces to $$|p|=\frac{h \nu}{c}=\frac{h}{\lambda}$$ or $$\nu =\frac{c}{\lambda}$$ But this does not limit the validity of the more general energy-momentum relation above.
