Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is $81 \mathrm{V}$ for the photoelectric emission experiment.

In the explanation of photoelectric effect, we assume one photon of frequency $v$ collides with an electron and transfers its energy. This leads to the equation for the maximum energy $E_{max}$ of the emitted electron as

$E_{max}=hv={\varphi }_0$ where ${\varphi }_0$ is the work function of the metal. If an electron absorbs $2$ photons (each of frequency $v$), what will be the maximum energy for the emitted electron?

In the explanation of photo electric effect, we assume one photon of frequency $v$ collides with an electron and transfers its energy. This leads to the equation for the maximum energy $E_{max}$ of the emitted electron as

$E_{max}=hv={\varphi }_0$ where ${\varphi }_0$ is the work function of the metal. If an electron absorbs $2$ photons (each of frequency $v$), what will be the maximum energy for the emitted electron? Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Consider figure for photoemission.

How would you reconcile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons.

The wavelength of light from the spectral emission line of sodium is $589 \mathrm{nm}$. Find the kinetic energy at which a neutron would have the same de Broglie wavelength.

An electron and a photon each have a wavelength of 1.00 nm. Find their momenta.