# The following graph shows t he variation of stopping potential V 0 with the frequency ν of the incident radiation for two photosensitive metals X and Y: (i) Which of the metals has larger threshold wavelength? Give reason. (ii) Explain giving reason, which metal gives out electrons, having larger kinetic energy, for the same wavelength of the incident radiation. (iii) If the distance between the light source and metal X is halved, how will the kinetic energy of electrons emitted from it change? Give reason Solution: (i) Let = Frequency of incident radiations of metal Y = Frequency of incident radiations of metal X  Therefore, metal X has larger threshold wavelength. (ii) Since the kinetic energy of the emitted electrons is directly proportional to the frequency of incident radiation, metal Y having larger incident frequency will have larger kinetic energy. ∴ Metal Y has larger kinetic energy. (iii) Kinetic energy of the emitted photoelectrons is independent of the intensity of the incident light. Hence, kinetic energy of the emitted photoelectrons remains unchanged if the distance between the light source and metal X is halved. in this in (ii) part why work function.is not considered kindly reply fast... ??

The kinetic energy of the photoelectrons only depends upon the frequency of the incident radiation, not its wave function. Wave function only determines whether a particular electron will be ejected by the action of incoming energetic photon. So, work function is a constant for any material.

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in this part (ii) since the wave length of radiation is same maximum kinetic energy depend on the work function bcz          k.e(max)=eV=hv-(work function) since wave length is same the term hv is same for all so k.e(max) depend on (work function).

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