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

The graph shows the variation of de-Broglie wavelength $\lambda$ versus $\frac{1}{\sqrt{V}}$,where $V$ is the accelerating potential for four particles carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents the particle with the smallest mass?

Proton $(P)$ and electron $(e)$ will have the same de Broglie wavelength when the ratio of their momentum is (assume,$m_{p} = 1849 \, m_{e}$)

The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of $10^{-15} \;m$ or less. This structure was first probed in early $1970s$ using high energy electron beams produced by a linear accelerator at Stanford,$USA$. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron $= 0.511 \;MeV$.)

The de Broglie wavelengths of an electron $(e)$,proton $(p)$,neutron $(n)$,and $\alpha$-particle are $\lambda_e, \lambda_p, \lambda_n$,and $\lambda_\alpha$ respectively. All have the same kinetic energy of $1 \ MeV$. Which of the following represents the correct increasing order of their wavelengths?

If the kinetic energy of the particle is increased by $16$ times,the percentage change in the de Broglie wavelength of the particle is ............$\%$