Q.(i) In the explanation of photoelectric effect, we assume one photon of frequency $\nu$ collides with an electron and transfers its energy. This leads to the equation for the maximum energy $E_{max}$ of the emitted electron as $E_{max} = h\nu - \phi_0$, where $\phi_0$ is the work function of the metal. If an electron absorbs 2 photons (each of frequency $\nu$) what will be the maximum energy for the emitted electron?

The key idea is the photoelectric effect and the assumption of a one-photon–one-electron interaction.

(i) If a single electron absorbs two photons simultaneously, it gains the energy of both photons. The maximum kinetic energy of the emitted electron then becomes:

This is because the electron must still overcome the work function $\phi_0$ once to escape the metal. …

The core idea: energy conservation in the photoelectric effect

The photoelectric equation $E_{\text{max}} = h\nu - \phi_0$ comes from a simple energy‑balance idea: one photon gives all its energy $h\nu$ to one electron. The electron uses $\phi_0$ of that energy to escape the metal, and the rest becomes kinetic energy. If the electron could absorb two photons at once, the energy input would double — but the work function $\phi_0$ stays the same (it’s a property of the metal, not of the light).

So the question is really: what happens to the energy balance when the electron gets two photons instead of one?

(i) Maximum energy with two‑photon absorption

1. Energy input Each photon carries energy $h\nu$. If an electron absorbs two such photons simultaneously, the total energy it receives is

2. Energy cost to escape

   No matter how many photons are absorbed, the electron still needs to overcome the same work function $\phi_0$ to leave the metal. That energy is lost from the absorbed energy.

3. Remaining kinetic energy

   The maximum kinetic energy occurs when the electron uses the absorbed energy as efficiently as possible — i.e., it loses exactly $\phi_0$ and converts the rest to motion. So

4. Comparison with one‑photon case For one photon: $E_{\text{max}}^{(1)} = h\nu - \phi_0$. For two photons: $E_{\text{max}}^{(2)} = 2h\nu - \phi_0$. The difference is $h\nu$ — an extra photon’s worth of energy.

(ii) Why two‑photon absorption is ignored in stopping potential discussions

The stopping potential $V_0$ is defined by $eV_0 = E_{\text{max}}$. In standard experiments, we measure $V_0$ and use the one‑photon equation to find $h$ or $\phi_0$. Two‑photon absorption would give a higher $E_{\text{max}}$, so why don’t we see it?

1. Probability is vanishingly small at ordinary intensities …

Method: Extending the Photoelectric Equation to Multi-Photon Absorption, and Judging When a Process Is Negligible

Two general skills in one question: modifying a standard formula when an assumption behind it is relaxed, and reasoning about whether a physically possible process actually matters under realistic conditions.

Steps

Step 1: Identify exactly which assumption is being relaxed

The standard equation $E_{\max} = h\nu - \phi_0$ rests on the assumption "one photon transfers its energy to one electron." When a question changes that assumption (here: the electron absorbs $n$ photons instead of one), go back to the energy-conservation argument the formula came from, rather than trying to pattern-match a new formula.

Step 2: Re-apply energy conservation with the new input energy

The total energy delivered to the electron is now $n$ photons' worth, $n h\nu$. The escape cost is still paid once — the work function is a property of the metal the electron must overcome exactly once to leave, regardless of how many photons supplied the energy:

For $n=2$ specifically: $E_{\max} = 2h\nu - \phi_0$.

Step 3: For the "why isn't this considered" half — compare probabilities, not possibilities …