$A$ proton and an electron have the same de Broglie wavelength. If $K_p$ and $K_e$ are the kinetic energies of the proton and electron respectively,then choose the correct relation:

Let $E_e$ and $E_p$ represent the kinetic energy of an electron and a photon,respectively. If the de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron,then find the ratio $E_p / E_e$. (Given: speed of electron $v = c / 100$,where $c$ is the velocity of light).

$A$ proton moving with one-tenth of the velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having a certain kinetic energy has the same de Broglie wavelength $\lambda$. The ratio of the kinetic energy of the proton to that of the alpha particle is:

$A$ proton and an $\alpha$-particle are accelerated through a potential difference of $100\, V$. The ratio of the wavelength associated with the proton to that associated with an $\alpha$-particle is

An electron with speed $v$ and a photon with speed $c$ have the same $de-Broglie$ wavelength. If the kinetic energy and momentum of the electron are $E_{e}$ and $p_{e}$ and that of the photon are $E_{ph}$ and $p_{ph}$ respectively,which of the following is correct?

The ratio of the de-Broglie wavelengths of a proton and an electron having the same kinetic energy is: (Assume $m_p = 1849 \times m_e$)