The kinetic energy of an electron is $5 \ eV$. Calculate the de-Broglie wavelength associated with it in $\mathring{A}$. $(h = 6.6 \times 10^{-34} \ J \cdot s, m_e = 9.1 \times 10^{-31} \ kg)$

An $\alpha$-particle moves in a circular path of radius $1 \ cm$ in a uniform magnetic field of $0.125 \ T$. The de Broglie wavelength associated with the $\alpha$-particle is

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 $P_m$)

What is the de Broglie wavelength of an electron accelerated through $80 \ V$ in $\mathring A$?

The de-Broglie wavelength of an electron with kinetic energy of $320 eV$ is (Take $h = 6.0 \times 10^{-34} \text{ SI unit}$, mass of electron $m_{e} = 9.0 \times 10^{-31} \text{ kg}$, charge of an electron $e = 1.6 \times 10^{-19} \text{ C}$). (in $pm$)

The de Broglie wavelengths associated with a proton and an electron are in the ratio $2: 1$. Their stopping potentials are approximately in the ratio of