The de Broglie wavelength associated with a neutron at temperature $T$ is given by: $(E = kT)$

$A$ wave is associated with matter:

The velocity of a particle $A$ is $3$ times the velocity of a proton. If the ratio of the de Broglie wavelengths of the particle $A$ and the proton is $3:2$,the mass of the particle $A$ is (where $m_{p}$ is the mass of the proton).

The proton and $\alpha$-particle are accelerated through the same potential difference. Then the ratio of the de-Broglie wavelength of proton and $\alpha$-particle is (mass of $\alpha$-particle is $4$ times mass of proton, charge of $\alpha$-particle is $2$ times charge of proton).

When energy is added to an electron,its de Broglie wavelength decreases from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$. The added energy is:

Two particles move at right angles to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles undergo a perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle is given by