The de Broglie wavelengths of a proton and an $\alpha$ particle are $\lambda$ and $2\lambda$ respectively. The ratio of the velocities of proton and $\alpha$ particle will be:

Crystal diffraction experiments can be performed using $X-$rays,or electrons accelerated through an appropriate voltage. Which probe has greater energy? (For quantitative comparison,take the wavelength of the probe equal to $1\,\mathring{A}$,which is of the order of inter-atomic spacing in the lattice) $(m_{e}=9.11 \times 10^{-31}\,kg)$

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

Find the ratio of the de Broglie wavelengths of a proton and an $\alpha$-particle accelerated through the same potential difference.

The de-Broglie wavelength of an electron is the same as that of a $50 \ keV$ $X$-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \ MeV$ ).

$A$ particle of mass $9.1 \times 10^{-31} \, \text{kg}$ travels in a medium with a speed of $10^{6} \, \text{m/s}$ and a photon of radiation with linear momentum $10^{-27} \, \text{kg} \cdot \text{m/s}$ travels in vacuum. The wavelength of the photon is $....$ times the wavelength of the particle.