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

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

An electron of mass $m$ has de-Broglie wavelength $\lambda$ when accelerated through potential difference $V$. When a proton of mass $M$ is accelerated through a potential difference of $9V$,the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage).

The de-Broglie wavelength of an electron with kinetic energy of $320 eV$ is (Take $h = 6.0 \times 10^{-34} \text{ SI unit}$, mass of electron $m_{e} = 9.0 \times 10^{-31} \text{ kg}$, charge of an electron $e = 1.6 \times 10^{-19} \text{ C}$). (in $pm$)

The de Broglie wavelength associated with a neutron at temperature $T$ is given by: $(E = kT)$

$A$ photosensitive metallic surface emits electrons when $X$-rays of wavelength $\lambda$ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface,$m$ is mass of the electron,$h$ is Planck's constant,$c$ is the velocity of light).