The kinetic energy of an electron is increased by $2$ times,then the de-Broglie wavelength associated with it changes by a factor.

Which of the following has the largest de Broglie wavelength?

Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\left(\frac{\lambda_1}{\lambda_2}\right) \propto E^{n}$,then the value of $n$ is:

The de-Broglie wavelength of an electron (mass $= 1 \times 10^{-30} \ kg$, charge $= 1.6 \times 10^{-19} \ C$) with a kinetic energy of $200 \ eV$ is (Planck's constant $= 6.6 \times 10^{-34} \ J \cdot s$):

$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

The de Broglie wavelength of an electron moving with a velocity $1.5 \times 10^8 \, m/s$ is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is