In photoelectric effect, when frequency of light is increased keeping the intensity same then what effects will be observed on saturation current and stopping potential.

Explain with regards to failure of wave theory and success of photon theory.

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## Comments

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TopNewestIt's important that what else is tried is that the frequency is kept the same but the intensity is increased. Let the two methods be listed, and results described:

1) Intensity same, frequency increased: (many) metals emit electrons

2) Frequency same, intensity increased: Same metals do not emit electrons

The wave theory suggests that both 1) and 2) result in pumping increasing energy into the metal, so one would expect that eventually electrons will be emitted. But since that only happens with 1), then whether or not metals emit electrons depends on the frequency, or threshold frequency. This suggests a different conceptual model for light, since the wave model doesn't explain it. The theory of quanta was developed from this, leading to photons as particles rather than waves. Eventually, the two concepts are combined in quantum theory.

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If freq is increased keeping the intensity same ... Then energy of individual photons will increase to keep intensity same then number of photons per sec per unit area should decrease thus photoectric current should decrease as well.. But results are contradictory.. So the question is how photon theory is explaining the observation...

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It is the energy of EACH photon that can cause a metal atom to emit an electron. If the energy of each photon is below a certain threshold, then it doesn't matter how many photons bombard the metal atom, an electron will not be emitted.

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1) The probability of electrons being emitted goes up. This does not mean that the number emitted steadily ramps up indefinitely. It levels off to where that the probability of an electron being emitted for each photon used goes up to near 100%, one-for-one.

2) The extra energy in higher frequency photons used will result in more energic electrons. This is what will ramp up [almost] indefinitely. The energy in the photons used is transferred into the energy of the electrons emitted, individually.

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do you agree with this??

PS: Efficiency of photoelectric effect is very low thus by increasing the frequency the efficiency doesn't get readily affected.!!

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Now, even when an individual photon does have the threshold frequency (or energy), that does not mean that when it strikes a metal atom, an electron "will" always emit an electron. It's still probabilistic. It's just that for photons with less than threshold energy, the probability is 0. With higher frequency photons, the probability approaches 100%

Intensity denotes the number of photons flowing per unit time. That is not the same as total energy of photons flowing per unit of time. So, for example, we can have the same intensity of red or blue light, but the latter will carry more energy.

Likewise, for photons of the same frequency, increasing intensity of light will mean greater flow of energy, simply because there are more "packets of energy" flowing per unit time.

"Photocurrent" refers to the electric flow in a photosensitive device. So, it's quite possible that for the same intensity incident red or blue light, the photocurrent is 0 with the former, and not-0 with the latter.

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