The ratio of kinetic energies of a proton and an $\alpha$-particle is $16:1$. What is the ratio of their associated de Broglie wavelengths?

The ratio of the accelerating potentials required to accelerate $(i)$ an $\alpha$-particle and (ii) a proton to have the same de Broglie wavelength associated with them is (mass of $\alpha$-particle = $6.4 \times 10^{-27} \text{ kg}$, mass of proton = $1.6 \times 10^{-27} \text{ kg}$)

An electron microscope uses electrons of $40 \ keV$. The de Broglie wavelength associated with these electrons is approximately:

If the kinetic energy of a free electron doubles,its de-Broglie wavelength $\lambda$ changes by a factor of:

The de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda$. What will be its wavelength at $927^{\circ} C$?

$A$ particle of mass $8 \mu g$ in motion collides with another stationary particle of mass $4 \mu g$. If the collision is perfectly elastic and one-dimensional, the ratio of their de Broglie wavelengths after collision is (in $2 : 1$)