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In a photoelectric effect experiment, photons of energy 5 eV are incident on the photo-cathode of work function 3 eV . For photon intensity I _ A =10¹⁵ ~m ⁻² ~s ⁻¹ , saturation current of 4.0 ~A is obtained. Sketch the variation of photocurrent i_p against the anode voltage V _ a for photon intensity I _ A (curve A ) and I_B=2 10¹⁵ ~m ⁻² ~s ⁻¹ (curve B) will be:

Correct answer

0

Step-by-step solution

Stopping potential, V =( h v- W ) / e =2 ~V In each case, the stopping potential is 2 V as energy of incident photon is the same. As the photon intensity is doubled, the saturation current is also doubled. The variation of i_p against V_a is sketched in the figure

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