Relativistic change of mass and De-broglie's equation From De Broglie's equation $$\lambda=\frac{h}{mv}$$
We can see that $\lambda$ depends inversely on $m$
My question is:
If the $m$ in the equation indicates the kinetic mass of the body then from this relation$$m=\frac{m_0}{√(1-v^2/c^2)}$$ $m$ would increase with $v$. Thus the $\lambda$ value should change doubly...first for the rise of $v$ and second for the rise of $m$
Or is the $m$ in De Broglie's equation representing rest mass, meaning that I am altogether wrong.
I am a layman in Physics and I stumbled on this question while understanding electron microscopy
 A: Let us restrict our studies to one spatial dimension plus time, coordinates $(t,x).$
In the frame where the particle is in rest, consider a field $\phi$ that oscillates with a frequency proportional to the rest mass/rest energy:
$\phi = \sin \omega_0 t$
where $\omega_0 = 2\pi f_0 = 2\pi \, m_0 c^2/h = m_0 c^2/\hbar.$
Now we introduce a relativistic inertial frame $(t',x')$ in which the particle moves with speed $v$ in the positive $x$ direction. The Lorentz transformation between these frames is
$$x' = \gamma (x + tv), \quad t' = \gamma (t + xv/c^2)$$
$$x = \gamma (x' - t'v), \quad t = \gamma (t' - x'v/c^2)$$
where $\gamma = 1/\sqrt{1-v^2/c^2}.$
Thus,
$$
\phi 
= \sin \frac{m_0 c^2}{\hbar}t 
= \sin \frac{m_0 c^2}{\hbar} \gamma(t' - x'v/c^2)
= \sin \frac{\gamma m_0 c^2}{\hbar} (t' - x'v/c^2)
= \sin \frac{m c^2}{\hbar} (t' - x'v/c^2) \\
= \sin \frac{m c^2 t' - mv x'}{\hbar}
= \sin \frac{E t' - p x'}{\hbar}
= \sin (\omega t' - k x')
$$
The wavelength is then given by
$$
\lambda = \frac{2\pi}{k} = \frac{2\pi}{p/\hbar} = \frac{2\pi \, \hbar}{p} = \frac{h}{p}.
$$
A: In case you want a less mathematical answer:
what you say is not wrong but a rather upside-down way of thinking of it.
When de Broglie first came up with formula $ \lambda = \frac h p $ in 1923 he was motivated by the way quantities transform under Einstein's relativity theory.
Einstein had previously established two principles:


*

*Energy is directly proportional to frequency. $ (E = hf) $

*Energy and momentum $(E,p)$ naturally transform together when you change your reference frame


So the only way to make sense of a relativistic quantum matter wave is if you pair up $(f,\lambda)$ in a similar way, meaning that wavelength and momentum must be related by the same constant h. $ ( p = \frac h  \lambda ) $
So the complexity eluded to in your question does not come from quantum theory, but rather Einstein's original relativity theory: the relation between momentum and velocity is not a simply linear relation $( p = mv) $ but a more complex one where $v$ appears twice:
$$ p = \frac {m v} { \sqrt {1 - \frac {v^2} {c^s} }} $$ 
[$m$ denotes rest mass here] 
Reference: Weinberg, The Quantum Theory of Fields, Volume 1.
