$A$ proton and an $\alpha$-particle are accelerated through a potential difference of $100\, V$. The ratio of the wavelength associated with the proton to that associated with an $\alpha$-particle is

The magnitude of the de-Broglie wavelength $(\lambda)$ of electron $(e)$,proton $(p)$,neutron $(n)$ and $\alpha-$ particle $(\alpha)$ all having the same energy of $1\,MeV$,in the increasing order will follow the sequence

Electrons used in an electron microscope are accelerated by a voltage of $25 \ kV$. If the voltage is increased to $100 \ kV$,then the de-Broglie wavelength associated with the electrons would

The de-Broglie wavelength of a particle of kinetic energy $K$ is $\lambda$. What will be the wavelength of the particle if its kinetic energy becomes $\frac{K}{4}$?

$A$ beam of electrons of energy $E$ scatters from a target having atomic spacing of $1 \, Å$. The first maximum intensity occurs at $\theta = 60^{\circ}$. Then $E$ (in $eV$) is: (Planck constant $h = 6.64 \times 10^{-34} \, Js$, $1 \, eV = 1.6 \times 10^{-19} \, J$, electron mass $m = 9.1 \times 10^{-31} \, kg$)

The relation between the wavelength of electromagnetic radiation $(\lambda)$ and the de Broglie wavelength of its quantum (photon) $(\lambda')$ is . . . . . . .