The wavelength of a photon is equal to the wavelength of an electron moving with a kinetic energy of $1.5 \, eV$. What is the energy of the photon?

An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field,the radius of the $n^{th}$ orbital will therefore be proportional to

If the kinetic energy of a particle is increased to $16$ times its previous value,the percentage change in the de-Broglie wavelength of the particle is

$A$ nucleus $A$,with a finite de Broglie wavelength $\lambda_A$,undergoes spontaneous fission into two nuclei $B$ and $C$ of equal mass. $B$ flies in the same direction as that of $A$,while $C$ flies in the opposite direction with a velocity equal to half of that of $B$. The de Broglie wavelengths $\lambda_B$ and $\lambda_C$ of $B$ and $C$ are respectively

If $E_p$ and $E_e$ represent the kinetic energy of a photon and an electron respectively. If the de-Broglie wavelength $\lambda_p$ of a photon is twice the de-Broglie wavelength $\lambda_e$ of an electron,then $E_e / E_p$ is (Speed of electron $= C/100$,where $C$ is the velocity of light).