What is the ratio of the de Broglie wavelengths of a deuteron and a proton accelerated through the same potential difference?

Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\left(\frac{\lambda_1}{\lambda_2}\right) \propto E^{n}$,then the value of $n$ is:

The graph shows the variation of de Broglie wavelength $(\lambda)$ versus $\frac{1}{\sqrt{V}}$,where '$V$' is the accelerating potential for four particles $A, B, C, D$ carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents a particle of the largest mass?

The ratio of kinetic energies of a proton and an $\alpha$-particle is $16:1$. What is the ratio of their associated de Broglie wavelengths?

The de-Broglie wavelength of a particle accelerated with $150 \ V$ potential is $10^{-10} \ m$. If it is accelerated by $600 \ V$ potential difference,its wavelength will be: ............... $\mathring{A}$

$A$ stationary mass $M$ splits into two fragments of masses $m_1$ and $m_2$. What is the ratio of their de Broglie wavelengths,${\lambda _1}/{\lambda _2}$?