$A$ ball of mass $0.12 \ kg$ is moving with a speed $20 \ m \ s^{-1}$. Then its de Broglie wavelength is . . . . . . . ( $h = 6.63 \times 10^{-34} \ J \ s$ )

If the momentum of an electron is changed by $P_m$ and the associated de Broglie wavelength changes by $0.50\ \%$, find the initial momentum of the electron. (in $P_m$)

The de-Broglie wavelength associated with a particle of mass $m$ moving with velocity $v$ is:

The kinetic energy of an electron and a proton is $10^{-32} \ J$. Then the relation between their de-Broglie wavelengths is

An electron accelerated through a potential of $10000 \ V$ from rest has a de-Broglie wavelength $\lambda$. What should be the accelerating potential, so that the wavelength is doubled (in $V$)?

$A$ proton moving with one-tenth of the velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having a certain kinetic energy has the same de Broglie wavelength $\lambda$. The ratio of the kinetic energy of the proton to that of the alpha particle is: