If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with electrons change?

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$,their de Broglie wavelength in the frame of reference attached to their centre of mass is

The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. The ratio of the kinetic energy of the electron to the energy of a photon is ($m=$ mass of electron,$c=$ velocity of light,$h=$ Planck's constant).

Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4V$. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by: (given $m_e$ is the mass of the electron and $m_p$ is the mass of the proton).

Electrons with de-Broglie wavelength $\lambda$ fall on the target in an $X$-ray tube. The cut-off wavelength of the emitted $X$-rays is

The work done to accelerate an electron from rest so that it can have a de Broglie wavelength of $6600 \text{ Å}$ is nearly (Planck's constant $= 6.6 \times 10^{-34} \text{ J s}$ and mass of electron $= 9 \times 10^{-31} \text{ kg}$)