$A$ ball weighing $25 \ g$ moves with a velocity of $6.6 \times 10^4 \ cm/sec$. Find the associated de Broglie wavelength.

If $a_0$ is the Bohr radius,then the de-Broglie wavelength of an electron revolving in the second excited state of the $H$ atom will be:

The wavelength of an electron and a neutron will become equal when the velocity of the electron is $x$ times the velocity of the neutron. The value of $x$ is (Nearest Integer). (Mass of electron is $9.1 \times 10^{-31} \ kg$ and mass of neutron is $1.6 \times 10^{-27} \ kg$)

The de-Broglie wavelength of a tennis ball of mass $60 \ g$ moving with a velocity of $10 \ ms^{-1}$ is approximately . . . . . . (Planck's constant $h = 6.63 \times 10^{-34} \ J \cdot s$).

$A$ proton accelerated through a potential $V$ has de-Broglie wavelength $\lambda$. Then the de-Broglie wavelength of an $\alpha$-particle,when accelerated through the same potential $V$ is :

The wavelength associated with a particle of mass $9.109 \times 10^{-31} \ kg$ moving with a velocity of $1.20 \times 10^5 \ m/s$ is ....... meters.