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enter image description here

image courtesy of http://www.learncbse.in/

Why is the saturation current the same, as you vary the frequency of incident light but keep its intensity constant?

If intensity is a measure of the energy of the incident beam, the product of the number of packets of light and the energy of each packet must be constant. So, if you increase the frequency, the energy of each packet must increase but the number of packets must decrease to keep the intensity constant. So, the decrease in the number of packets should correspond to the decrease in the number of electrons being ejected from the emitter plate. If the above discussion talks about taking place in unit time, shouldn't the saturation current decrease with increase in frequency, keeping intensity constant?

Note: Another thread I found on this website hinted at the intensity talked about here is the number of energy packets. But I'm not sure how this is a good way of defining intensity, especially when "comparing and contrasting" this quantized view with that of a classical wave.

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    $\begingroup$ You're confusing some of the terms which describe optical flux, as Swaroop's answer explains. $\endgroup$ – Carl Witthoft Mar 21 '16 at 14:09
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Well intensity refers to the number of photons. In that graph. The intensity of light is the same for all 3 frequencies. So when the potential is high enough such that all electrons are propelled to flow, that's the maximum. You cannot get any more electrons from anywhere. And current depends on the number of electrons. So when electrons are flowing, that's the maximum. Even though the frequency is different, the positive potential propels all the electrons to flow.

The intensity is a measure of number of photons. Not energy

What the graph is about is the regarding potential. You see the retarding potential is lower(higher magnitude, notice it's a negative sign, so -30 is lower than -20, but magnitude is higher) for a higher frequency because the negative potential is required to oppose the electrons. For a higher frequency, the energy of the electron is higher, because the photon has more energy and that photon can give more energy to the electron. So you need more energy to stop such a highly energetic electron, and that's why the extra energy is provided by the potential, where the opposing energy on the electron is -e*V.

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  • $\begingroup$ Thanks. Just a quick fact though. Isn't the intensity of a classical electromagnetic proportional to the spatial energy density? $\endgroup$ – IanDsouza Mar 23 '16 at 6:58
  • $\begingroup$ Sorry, I meant 'classical electromagnetic wave' $\endgroup$ – IanDsouza Mar 25 '16 at 3:41
  • $\begingroup$ Well intensity simply refers to the number of photons. So if you mean, density of photons, yes. @IanDsouza $\endgroup$ – Swaroop Joshi Mar 30 '16 at 9:14
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$I = neAv_{d}$

As, Swaroop explained n remains constant. e and A are also constant.

$v_{d} = \frac{eE\tau }{m}$

e, m and E are constant here, since applied voltage is constant.

$\tau$ It is usually constant for metal at given temperature.

So, I does not change.

Theoretical reason

As you can see varying frequency does not increase no. of electrons released. But, there is an increase in speed of electrons in vaccum tube. But, will speed of electrons increase in metal?

The answer is no. Speed of electrons in metal does not increase. This is because of second relation above. Or This is because as speed of electrons will increase, they will more probably strike with each other, counteracting to reduce speed of electron.

So, I hope you understand why Current does not increase.

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$I = neAv $

I doubt that it is appropriate to use as it is derived for an electron inside a metal, where we assume it to collide with lattice of metal and come to rest. While no such thing (I think) is going to happen when electrons are moving with constant acceleration, any collision may lead to a reduced velocity of one while the velocity of other electron will be increase (elastic collision); so according to me the best way to EXPLAIN this is to understand the intensity as a count of photons.

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