Of the following particles having the same kinetic energy,the one which has the largest de Broglie wavelength is

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength $\lambda$. When the potential is changed to $V_2$,its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_1}{V_2}\right)$ is

If the kinetic energy of an electron is increased four times,the wavelength of the de-Broglie wave associated with it would become

The ratio of the de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is $........$

The ratio of momentum of an electron and an alpha particle which are accelerated from rest by a potential difference of $100\,V$ is

$A$ photon and an electron (mass $m$) have the same energy $E$. The ratio $\left(\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}}\right)$ of their de Broglie wavelengths is: ($c$ is the speed of light)