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$(i)$ In the explanation of photoelectric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads to the equation for the maximum energy Emax of the emitted electron as $E_{max} = hf - \phi _0$ (where $\phi _0$ where do 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 ?
Solution
$(i)$ Suppose one electron absorbs two photons each having frequency $f$, out of which it spends $W$ amount of energy for its emission and remaining amount ( $2 h f-W$ ) is possessed by it in the form of kinetic energy after getting emitted. Hence,
$\mathrm{K}=2 h f-\mathrm{W}$
$\therefore \mathrm{K}_{\max }=2 h f-\mathrm{W}_{\min }$
$\therefore\mathrm{K}_{\max }=2 h f-\phi_{0} \quad\left(\because \mathrm{W}_{\min }=\phi_{0}\right)$
$(ii)$ If above assumption is true then according to work energy theorem,
$\text { taking } \mathrm{K}_{\max }=\mathrm{V}_{0} e, \quad \text { (where } \mathrm{V}_{0}=\text { stopping potential) }$
$\therefore 2 h f-\phi_{0}=\mathrm{V}_{0} e$
$\therefore \mathrm{V}_{0}=\left(\frac{2 h}{e}\right) f+\left(-\phi_{0}\right)$
Above equation is like equation of a straight line $y=m x+c$. Hence if we plot graph of $\mathrm{V}_{0} \rightarrow f$ experimentally then we should get its slope equal to $\left(\frac{2 h}{e}\right) .$ But experimentally we get this slope only $\left(\frac{h}{e}\right)$. Hence above assumption is proved to be wrong. That is why only such assumption is not considered in the discussion of stopping potential.
