Answer:
K.E. of the emitted electrons, \[{{K}_{\max }}=\frac{1}{2}m\upsilon _{\max }^{2}=hv-{{W}_{0}}\] Higher is the work function \[({{W}_{0}})\] of the metal, the lesser will be the K.E. of the emitted electrons. (a) If the intensity of the radiation is doubled, (i) stopping potential remains unchanged and (ii) the photoelectric current gets doubled. (b) If the frequency of the incident radiation is doubled, (i) the stopping potential gets doubled and (ii) the photoelectric current remains unaffected.
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