The kinetic energy of an electron having de-Broglie wavelength $\lambda$ is ($h=$ Planck's constant,$m=$ mass of electron).

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

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

$A$ particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813 \times 10^{-4}$. Calculate the particle's mass and identify the particle.

The energy that should be added to an electron,to reduce its de-Broglie wavelength from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$,will be

If a ball of mass $1 \ kg$ has a velocity of $1 \ m/s$,what is its de Broglie wavelength?