Question
The following graph shown the variation of stopping potential V0 with the frequency v of the incident radiation for two photosensitive metals P and Q:
(i) Explain which metal has smaller threshold wavelengths.
(ii) Explain, giving reason, which metal emits photoelectrons having smaller kinetic energy.
(iii) If the distance between the light source and metal P is doubled, how will the stopping potential change?
Solution
The graph shows the variation of stopping potential Vo with frequency .
(i) Suppose the frequency of incident radiations of metal P and Q be v0 and v0’ respectively.
From the graph,
Therefore, metal 'Q' has smaller wavelength.
(ii) As we know, energy of a photon is E = hv0
Energy is directly proportional to frequency. P has a lesser frequency.
Hence, metal 'P' has smaller kinetc energy.
(iii) If the distance between the light source and metal P is doubled, stopping potential remains unaffected because the value of stopping potential for a given metal surface does not depend on the intensity of the incident radiation. It depends on the frequency of incident radiation.
(i) Suppose the frequency of incident radiations of metal P and Q be v0 and v0’ respectively.
From the graph,
Therefore, metal 'Q' has smaller wavelength.
(ii) As we know, energy of a photon is E = hv0
Energy is directly proportional to frequency. P has a lesser frequency.
Hence, metal 'P' has smaller kinetc energy.
(iii) If the distance between the light source and metal P is doubled, stopping potential remains unaffected because the value of stopping potential for a given metal surface does not depend on the intensity of the incident radiation. It depends on the frequency of incident radiation.
