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

The energy of separation of an electron in a $H$ atom in an excited state is $3.4 \ eV$. The de-Broglie wavelength (in $\mathring{A}$) associated with the above electron is,if the radius of the first orbit of the $H$ atom is $0.53 \ \mathring{A}$.

If the kinetic energy of an electron is $2.8 \times 10^{-23} \ J$,then the de Broglie wavelength will be ....... $(m_e = 9.1 \times 10^{-31} \ kg)$.

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

Calculate the de-Broglie wavelength of an electron travelling at $1\%$ of the speed of light.

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?