(i) In the explanations of photoelectric effect, we assume one photon of frequency `nu` collides with an electron and transfer its energy. This leads to the equation for the maximum energy `E_(max)` of the emitted electron as `E_(max)=hnu-phi_(0)` Where `phi_(0)` is the work function of the metal. if an electron absorbs 2 photons (each of frequency v) what will be the maximum energy for the emitted electron?
(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Explanation

To solve the given question, we will break it down into two parts as specified.

### Part (i): Maximum Energy for the Emitted Electron Absorbing Two Photons

1. **Understanding the Photoelectric Effect**:
- In the photoelectric effect, when a photon of frequency \( \nu \) strikes an electron, it transfers energy to the electron. The maximum kinetic energy \( E_{max} \) of the emitted electron can be expressed as:
\[
E_{max} = h\nu - \phi_0
...