# Thought of the day _ 18_photoelectric effect

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.

Note by Rohit Gupta
2 years, 11 months ago

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It'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.

- 2 years, 11 months ago

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...

- 2 years, 11 months ago

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.

- 2 years, 11 months ago

My doubt is if the freq is above the threshold freq then on further increasing the freq keeping intensity same what will be the effect on photocurrent??

- 2 years, 11 months ago

Two things will happen, if the frequency is increased but the intensity remains the same:

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.

- 2 years, 11 months ago

If freq is increased then the energy of individual photons goes up thus to keep the intensity same number of photons should decrease thus photocurrent will also decrease...
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.!!

- 2 years, 11 months ago

What the photoelectric effect strongly suggests is the model of single particle-like photons hitting individual atoms, and knocking out electrons only if the photon hitting the atom has the threshold energy. Which is solely determined by its frequency. Not the number or "current" of photons hitting the metal.

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.

- 2 years, 11 months ago

Intensity of light is energy falling per sec per unit area or number of photons falling per sec per unit area...?? Which is crrect with respect to light in photoelectric effect..

- 2 years, 11 months ago

Okay, I correct myself. Radiant flux has energy as unit of measure. "Intensity" is the wrong word for what I'm thinking about.

- 2 years, 11 months ago