(i) In the explanation of photoelectric effect, we assume one photon of frequency v collides with an electron and transfer its energy.This leads to the equation for the maximum energy E_max of the emitted electron as \({ E }_{ max }=hv-{ \phi }_{ 0 }\)
Where \({ \phi }_{ 0 }\)is the work function of metal.If an electron absorbs two photons(each of frequency v) What will be the maximum energy for the emitted electron?
(ii) 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.

Draw a graph to show the variation of stopping potential with frequency of radiation incident on a metal plate. How can the value of Planck's constant be determined from this graph?

Find the ratio of the de-Broglie wavelength associated with protons accelerated through a potential of 128 V and a particles accelerated through a potential of 64 V.
To calculate the ratio of de-Baroglie wavelength of two particle find the ratio in terms of symbols and then put the given numerical values.

Calculate 
(i) momentum and 
(ii) de-Broglie wavelength of the electron accelerated through a potential difference of 56 V