The de Broglie wavelength of an electron moving with kinetic energy of $144 \;eV$ is nearly

If the de Broglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6 \times 10^{14} \, Hz,$ then the speed of the electron is equal to: (Speed of light $= 3 \times 10^8 \, m/s;$ Planck's constant $= 6.63 \times 10^{-34} \, J \cdot s;$ Mass of electron $= 9.1 \times 10^{-31} \, kg$)

Explain how,according to Born's probability interpretation,a wave having a single (unique) wavelength is extended all over space.

The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. The ratio of the kinetic energy of the electron to the energy of a photon is ($m=$ mass of electron,$c=$ velocity of light,$h=$ Planck's constant).

After absorbing a slowly moving neutron of mass $m_{N}$ (momentum $0$),a nucleus of mass $M$ breaks into two nuclei of masses $m_{1}$ and $5m_{1}$ $(6m_{1} = M + m_{N})$ respectively. If the de-Broglie wavelength of the nucleus with mass $m_{1}$ is $\lambda$,then the de-Broglie wavelength of the other nucleus will be:

An electron is moving through a field. It is moving $(i)$ opposite to an electric field and $(ii)$ perpendicular to a magnetic field as shown. For each situation, determine the de-Broglie wavelength of the electron: