The de Broglie equation describes the relationship between the wavelength associated with the motion of an electron and its:

The velocities of two particles $A$ and $B$ are $0.05 \ ms^{-1}$ and $0.02 \ ms^{-1}$ respectively. The mass of $B$ is five times the mass of $A$. The ratio of their de-Broglie's wavelength is

The atomic masses of $He$ and $Ne$ are $4$ and $20 \ amu$,respectively. The value of the de Broglie wavelength of $He$ gas at $-73^{\circ} C$ is $M$ times that of the de Broglie wavelength of $Ne$ at $727^{\circ} C$. $M$ is

If the wavelength and the distance traveled by an electron in $1 \ s$ are the same,calculate its velocity.

What accelerating potential must be imparted to a proton beam to give it an effective wavelength of $\lambda = 0.05 \ \mathring{A}$? (Given: $m_p = 1.672 \times 10^{-27} \ kg$,$h = 6.626 \times 10^{-34} \ J \cdot s$,$e = 1.602 \times 10^{-19} \ C$)

If the kinetic energy of an electron is increased four times,the wavelength of the de-Broglie wave associated with it would become