Photoelectric effect: why does stopping potential ignore intensity while saturation current doesn't? My brain says brighter = faster electrons

7 answers & replies

• [+][–]Nisha Verma12thSample· 12d ago

  Photon picture, which is what Einstein used. Light comes in packets (photons) of energy $h\nu$. One photon gives all its energy to one electron in a single event. Then $K_{max} = h\nu - \phi$. Increasing the intensity at the same frequency just means more photons per second, so more electrons are ejected per second (a larger photocurrent), but each electron still got the same $h\nu$. So $K_{max}$ and the stopping potential don't change with intensity.

  Increase the frequency instead and each photon carries more energy, so $K_{max}$ and $V_0$ go up.

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  • [+][–]Harsh Gupta12thSample· 12d agoAuthor

    ok so intensity = more photons, not stronger photons. that's the line. I'll write it as "bright = crowd, frequency = strength" and stop pretending I understood it before 😭

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• [+][–]Arjun Nair12thSample· 12d ago

  Your numerical is fine. Converting joules to eV by dividing by $1.6\times10^{-19}$ is correct. Shortcut worth memorising:

  $E\,(\text{eV}) = \frac{1240}{\lambda\,(\text{nm})}$

  so for 400 nm, $E = 3.1$ eV.

  $K_{max} = 3.1 - 2.0 = 1.1$ eV, and the stopping potential satisfies $eV_0 = K_{max}$, so $V_0 = 1.1$ V. The threshold wavelength of this metal would be $1240/2.0 = 620$ nm. Light of 700 nm wouldn't eject anything.

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• [+][–]Ananya Reddy11thSample· 12d ago

  I think the brighter the light the more energy each electron gets, so the K_max should also increase a little with intensity, but the graph in NCERT only shows the stopping potential for different frequencies?

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• [+][–]Tushar PawarDropperSample· 12d ago

  Ananya, no: for the same frequency, curves with different intensities in NCERT's $I$–$V$ graph have the same stopping potential (the curves cut the voltage axis at the same point) but different saturation currents. Intensity changes the height, not where it stops.

  Another graph worth knowing: $V_0$ vs $\nu$ is a straight line, since $eV_0 = h\nu - \phi$, i.e. $V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$. The slope is $h/e$ (the same for every metal, that's how Millikan measured $h$), and the intercept on the frequency axis is $\nu_0$. Different metals give parallel lines.

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  • [+][–]Ananya Reddy11thSample· 12d ago

    thank you, I see it in the graph now. same stopping point, different heights.

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• [+][–]Rahul DubeyDropperSample· 12d ago

  The classical wave theory failed on three counts: it predicted that (1) K should grow with intensity, (2) there is no threshold frequency, light of any frequency should do if bright enough, and (3) there should be a time lag while the electron collects energy. Experiments show none of those. Emission is practically instantaneous (about $10^{-9}$ s or less), and Einstein's photon picture explains all three.

  Also the $K_{max}$ in the formula is the maximum. Electrons from deeper inside lose some energy to collisions on the way out, so they emerge with less than $K_{max}$. The ones from the surface get $h\nu - \phi$.

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