An electron microscope uses which property of the electron?

The de-Broglie wavelength of a moving bus with speed $v$ is $\lambda$. Some passengers left the bus at a stoppage. Now, when the bus moves with twice its initial speed, its kinetic energy is found to be twice its initial value. What is the de-Broglie wavelength of the bus now?

If $E_p$ and $E_e$ represent the kinetic energy of a photon and an electron respectively. If the de-Broglie wavelength $\lambda_p$ of a photon is twice the de-Broglie wavelength $\lambda_e$ of an electron,then $E_e / E_p$ is (Speed of electron $= C/100$,where $C$ is the velocity of light).

Light of wavelength $500\, nm$ is incident on a metal with work function $2.28\, eV$. The de Broglie wavelength of the emitted electron is:

If the momentum of an electron is changed by $\Delta P,$ then the de Broglie wavelength associated with it changes by $0.5\%.$ The initial momentum of the electron will be

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is