When the kinetic energy of an electron is increased,the wavelength of the associated wave will

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

$A$ 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).

Which of the following particles has the shortest de-Broglie wavelength,assuming they have the same kinetic energy?

The wavelength of a de-Broglie wave is $2 \mu m$. Calculate its momentum. (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$)

$A$ particle of mass $8 \mu g$ in motion collides with another stationary particle of mass $4 \mu g$. If the collision is perfectly elastic and one-dimensional, the ratio of their de Broglie wavelengths after collision is (in $2 : 1$)