What is the theoretical formula for the de-Broglie wavelength of a particle?

If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor

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 de-Broglie wavelength associated with an electron,accelerated through a potential difference of $121 \ V$ is about:
[Take Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$,mass of electron $m = 9 \times 10^{-31} \ kg$,charge of electron $e = 1.6 \times 10^{-19} \ C$] (in $nm$)

$A$ nucleus of mass $M$ at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2M'}{3}$ (where $M' < M$). The ratio of the de Broglie wavelengths of the two parts will be:

$A$ particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$,having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles,$\lambda_1 / \lambda_2$ is