The de Broglie wavelength $(\lambda)$ depends on mass '$m$' and kinetic energy '$E$' according to which formula?

The ratio of de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is

$A$ 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?

An $\alpha$-particle and a deuteron are moving with velocities $v$ and $2v$ respectively. What will be the ratio of their de Broglie wavelengths?

Electrons used in an electron microscope are accelerated by a voltage of $25 \ kV$. If the voltage is increased to $100 \ kV$,then the de-Broglie wavelength associated with the electrons would

$A$ photosensitive metallic surface emits electrons when $X$-rays of wavelength $\lambda$ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface,$m$ is mass of the electron,$h$ is Planck's constant,$c$ is the velocity of light).