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

The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two jars kept separately at temperatures of $27\,^{\circ}\text{C}$ and $127\,^{\circ}\text{C}$ respectively is

When an electron is accelerated through a potential $V$,the de-Broglie wavelength associated with it is $4 \lambda$. When the accelerating potential is increased to $4V$,its wavelength will be

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 \,nm$ to $0.5 \,nm$ is

$A$ proton and an $\alpha$-particle are accelerated using the same potential difference. How are the de-Broglie wavelengths $\lambda_p$ and $\lambda_{\alpha}$ related to each other?

The potential difference $V$ required for accelerating an electron to have the de-Broglie wavelength of $1 \text{ Å}$ is (in $\text{ V}$)