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When a monochromatic point source of light is at a distance of 0.2m from a photocell, the cut-off voltage and saturation current are 0.6V and 18mA respectively.If the same source is placed 0.6m away from the photoelectric cell, thenthe stopping potential will be 0.2 V the stopping potential will be 0.6V the saturation potential will be 6mAanswer is b. I could not solve the question. Please help to do it.

Chandra Chatterjee , 10 Years ago

Grade 12

2 Answers

Mallikarjun Maram

Last Activity: 10 Years ago

When the distance of the source is increased, the intensity of light falling on the photcell is reduced. It does not reduce the energy of individual photons that are incident on the cell. Hence maximum KE of emmitted photo electrons will be same as before. Stopping potential and maximum KE are related by KE_max = eV. Since KE_max is not changing, stopping potential will not change.

However, as the intensity is reduced, number of photoelectrons emmitted will be reduced and hence saturation current will reduce. However it will not be 6mA (option C), as the intensity varies inversely with square of the distance.

Approved

Rishi Sharma

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

As the distance of the source is increased, the intensity of light falling on the photcell is reduced.
It does not reduce the energy of individual photons that are incident on the cell.
Hence maximum KE of emmitted photo electrons will be same as before.
Stopping potential and maximum KE are related by
KEmax = eV.
Since KEmax is not changing, stopping potential will not change.
However, as the intensity is reduced, number of photoelectrons emmitted will be reduced and hence saturation current will reduce.
However it will not be 6mA (option C), as the intensity varies inversely with square of the distance.

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