If $\lambda$ and $K$ are the de Broglie wavelength and kinetic energy,respectively,of a particle with constant mass,the correct graphical representation for the particle will be:

$A$ subatomic particle of mass $10^{-30} \ kg$ is moving with a velocity $2.21 \times 10^6 \ m/s$. Under the matter wave consideration,the particle will behave closely like . . . . . . . $(h = 6.63 \times 10^{-34} \ J \cdot s)$

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

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

$A$ particle is travelling $4$ times as fast as an electron. Assuming the ratio of the de-Broglie wavelength of the particle to that of the electron is $2:1$,the mass of the particle is:

The kinetic energy of a free electron increases to $3$ times the previous kinetic energy $(K.E.)$. The ratio of the new de-Broglie wavelength to the previous de-Broglie wavelength is: