Question

In the explanation of photo electric 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 $E_{max}$ of the emitted electron as $E_{max}=hv-{\varphi }_{\theta }$ where ${\varphi }_{\theta }$ is the work function of the metal. If an electron absorbs $2$ photons (each of frequency $v$)

$A$: what will be the maximum energy for the emitted electron?


$B$: Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Answer 1

$A$:

Given, frequency of each electron $=v$

Maximum energy of emitted electron,
$E_{max}=hv-{\varphi }_{\theta }$
As the electron absorb $2$ photons.

Frequency of emitted electron, $v^{{}^{\prime }}=2v$

Therefore, maximum energy for emitted electron is,
$E_{max}=h{v}^{{}^{\prime }}-{\varphi }_0$
$E_{max}=2hv-{\varphi }_0$
Final Answer: $2hv-{\varphi }_0$

$B$:

The probability of absorbing two photons by the same electron is very low. Hence such emission will be negligible.