The potential difference $V$ required for accelerating an electron to have the de-Broglie wavelength of $1 \text{ Å}$ is (in $\text{ V}$)

$A$ 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:

Of the following particles having the same kinetic energy,the one which has the largest de Broglie wavelength is

If the de Broglie wavelengths associated with a proton and an $\alpha$-particle are equal,then the ratio of velocities of the proton and the $\alpha$-particle will be

Two particles of equal masses are moving with equal speeds at an angle $60^o$. The de-Broglie wavelength of these particles is $\lambda$. Find the de-Broglie wavelength of the particles in the frame of the centre of mass of the particles.

An electron and a photon each have a wavelength of $1.00 \; nm$. Find
$(a)$ their momenta,
$(b)$ the energy of the photon,and
$(c)$ the kinetic energy of the electron.