If we consider an electron and a photon of the same de-Broglie wavelength,then they will have the same:

$A$ 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

The de Broglie wavelength of a charged particle accelerated through a potential difference $V$ is $\lambda$. If the potential difference is increased by $21 \%$,the de Broglie wavelength of the charged particle is

According to the de-Broglie hypothesis, the ratio of the wavelength of a photon and that of an electron having the same energy $E$ is $(m = \text{mass of electron}, c = \text{velocity of light})$

An electron beam used in an electron microscope,when accelerated by a voltage of $20\,kV$,has a de-Broglie wavelength of $\lambda_0$. If the voltage is increased to $40\,kV$,then the de-Broglie wavelength associated with the electron beam would be:

If the kinetic energy of a moving particle is $E$,then the De Broglie wavelength is