In a photoelectric experiment anode potential is plotted against plate current
(IIT-JEE-2004)
In a photoelectric experiment anode potential is plotted against plate current
(IIT-JEE-2004)
The correct option is A A and B will have different intensities while B and C will have different frequencies.
i vs v graph of a photoelectric experiment looks like the above. Now imagine if in the same experiment if I brought the light source closer to the plate. I have not changed frequency but I have increased intensity by doing that. I α nhv
I ↑ v is constant so n will ↑
If number of photons falling on a plate has increased then number of photo electrons coming out will also increase. Hence the current will have to increase.
Also since the frequency is same (same source) then the energy of each photon is going to be same. Hence the energy of highest energy electron won’t change right. So the stopping potential should not change too.
Now if I increase the frequency of each packet then I have increased the energy of each packet. This means that now the energy of the highest energy electron is also more than before. This means to stop this electron we will need a greater stopping potential.
NCERT states if in 2 experiments, Lights intensity falling on plate is same irrespective of the fact whether they are of same colour of or not, the saturation current will remain same.
Going by the above option (a) is the correct choice
A and B will have different intensities while B and C will have different frequencies.
i vs v graph of a photoelectric experiment looks like the above. Now imagine if in the same experiment if I brought the light source closer to the plate. I have not changed frequency but I have increased intensity by doing that. I α nhv
I ↑ v is constant so n will ↑
If number of photons falling on a plate has increased then number of photo electrons coming out will also increase. Hence the current will have to increase.
Also since the frequency is same (same source) then the energy of each photon is going to be same. Hence the energy of highest energy electron won’t change right. So the stopping potential should not change too.
Now if I increase the frequency of each packet then I have increased the energy of each packet. This means that now the energy of the highest energy electron is also more than before. This means to stop this electron we will need a greater stopping potential.
NCERT states if in 2 experiments, Lights intensity falling on plate is same irrespective of the fact whether they are of same colour of or not, the saturation current will remain same.
Going by the above option (a) is the correct choice
