An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75 \%$ of the initial value?

If the de Broglie wavelengths of an electron and a proton are equal,then the kinetic energy of the electron is ........

Two particles move at right angles to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles undergo a perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle is given by

De Broglie waves are associated with moving particles. These particles can be .......

$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)