$A$ proton and an $\alpha$-particle are accelerated from rest to the same kinetic energy. The ratio of their de Broglie wavelengths $\lambda_{p} : \lambda_{\alpha}$ is:

If the de Broglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6 \times 10^{14} \, Hz,$ then the speed of the electron is equal to: (Speed of light $= 3 \times 10^8 \, m/s;$ Planck's constant $= 6.63 \times 10^{-34} \, J \cdot s;$ Mass of electron $= 9.1 \times 10^{-31} \, kg$)

The de Broglie wavelength of an electron of kinetic energy $9 \ eV$ is (take $h=4 \times 10^{-15} \ eV \cdot s$,$c=3 \times 10^{10} \ cm/s$ and the mass $m_e$ of electron as $m_e c^2=0.5 \ MeV$)

The ratio of de Broglie wavelengths associated with thermal neutrons at temperatures $127^{\circ} C$ and $352^{\circ} C$ is

An electron beam,when accelerated by a voltage of $10 \ kV$,has a de-Broglie wavelength of $\lambda$. If the voltage is increased to $20 \ kV$,then the de-Broglie wavelength associated with the electron beam would be:

$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of