If the kinetic energy of a free electron is doubled,by what factor does its de Broglie wavelength change?

$A$ particle $P$ is formed due to a completely inelastic collision of particles $x$ and $y$ having de-Broglie wavelengths $\lambda_x$ and $\lambda_y$ respectively. If $x$ and $y$ were moving in opposite directions,then the de-Broglie wavelength of $P$ is

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

The ratio of de Broglie wavelengths associated with thermal neutrons at temperatures $127^{\circ} C$ and $352^{\circ} C$ is

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

The de Broglie wavelength of a proton accelerated by a potential difference of $100 \ V$ is $\lambda_0$. If an alpha particle is accelerated by the same potential difference,its de Broglie wavelength will be: