An electron,an $\alpha$-particle,and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?

Particle $A$ of mass $m_{A} = \frac{m}{2}$ moving along the $x$-axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B} = \frac{m}{3}$. If both particles move along the $x$-axis after the collision,the change $\Delta \lambda$ in the de-Broglie wavelength of particle $A$,in terms of its de-Broglie wavelength $(\lambda_{0})$ before the collision is:

If the momentum of an electron changes by $P$,then the de-Broglie wavelength associated with it changes by $5 \%$. The initial momentum of the electron is: (in $P$)

$A$ photon and an electron are given the same energy $(10^{-20} \ J)$. If the wavelengths associated with the photon and the electron are $\lambda_{Ph}$ and $\lambda_{el}$ respectively,then which of the following statements is correct?

The de-Broglie wavelength associated with a particle of mass $m$ moving with velocity $v$ is:

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