The idea of matter waves was given by

$A$ proton and a photon both have the same energy of $E = 100 \,keV$. If the de Broglie wavelength of the proton is $\lambda_1$ and that of the photon is $\lambda_2$,then $\lambda_1/\lambda_2$ is proportional to:

Calculate the $(a)$ momentum,and $(b)$ de Broglie wavelength of the electrons accelerated through a potential difference of $56 \; V$.

What is the experimental value of the de-Broglie wavelength for an electron accelerated through a potential difference of $V$ volts?

The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of $3 \ km/s$ will be ............ $\mathring{A}$.

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is ($c$ = velocity of light in air).