The de-Broglie wavelength associated with a particle of mass $m$ moving with velocity $v$ 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

The ratio of the de-Broglie wavelength of molecules of hydrogen $(H_2)$ and helium $(He)$ which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is:

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?

The de Broglie wavelength of an electron accelerated between two plates having a potential difference of $900 \ V$ is nearly. (in $nm$)

The ratio of wavelengths of a proton and a deuteron accelerated by potentials $V_{p}$ and $V_{d}$ is $1 : \sqrt{2}$. Then,the ratio of $V_{p}$ to $V_{d}$ will be: