The relationship between the momentum $(P)$ and the wavelength $(\lambda)$ of a particle is given by:

An electron has kinetic energy $2.8 \times 10^{-23} \ J$. The de-Broglie wavelength will be nearly $(m_e = 9.1 \times 10^{-31} \ kg)$.

What should be the ratio of velocities of $CH_4$ and $O_2$ molecules so that they are associated with de Broglie waves of equal wavelength?

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.

Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope, often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is $1.6 \times 10^{6} \,ms^{-1}$, calculate the de Broglie wavelength associated with this electron.

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