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

The de Broglie wavelength for a deuteron can be given by:

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

$A$ long pipe is filled with small granules of nickel. The interatomic spacing for nickel is $d$. $A$ beam of electrons having different energies is incident on the left flat face of the pipe. Mass of electron is $m$. The beam of electrons coming out of the right flat face has energy:

The de Broglie wavelength associated with an electron accelerated through a potential difference of $\frac{200}{3} \,V$ is nearly (in $Å$)

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same