Photoelectric effect , Superposition of sine waves

[Terry Bing]

Homework Statement

In a photoelectric effect experiment, a monochromatic plane wave of light falls on a metal plate. The electric field in the light wave at a point near the plate varies according to $E=E_0 \cos (\omega t)$. This results in a saturation current of 6 μA. If instead, the light wave was governed by $E=E_0 (1+\cos (\omega t)) \cos (\omega t)$, what would the saturation current be?

Homework Equations

The saturation current $I_s \propto n_p$ where $n_p$ is the number of incident photons per second and $n_p \propto {E_0}^2$.

The Attempt at a Solution

The given wave can be written as a sum of 3 sinusoids of frequency $\omega$, $2 \omega$ and zero.
$E=E_0 \cos (\omega t)+ \frac{E_0}{2} \cos (2 \omega t)+\frac{E_0}{2}$
Since frequence $\omega$ can knock out photoelectrons, so can $2 \omega$. The third term doesn't eject pholoelectrons.
I found the number of photons per second in the two waves (this is the part I suspect is wrong)
$$n_{p,\omega} \propto {E_0}^2$$
$$n_{p,2 \omega} \propto \left( \frac{E_0}{2}\right) ^2$$
and total photons incident per second would be
$$n_p' = n_{p,\omega} + n_{p,2 \omega} \propto \frac{5}{4} {E_0}^2$$
The saturation current in this case
$$I_s' \propto n_p' \propto \frac{5}{4} {E_0}^2$$
$$\implies I_s'=\frac{5}{4} I_s= 7.5 μA$$
However, the answer given at the back of the book is 6.75 μA which is like 9/8 I_s. What am I doing wrong

 

[kuruman]

The saturation current $I_s \propto n_p$ where $n_p$ is the number of incident photons per second and $n_p \propto {E_0}^2 .$

I think you should pay closer attention to the constant of proportionality and how it is obtained. Is it the same in the two situations?

 

[Terry Bing]

I think you should pay closer attention to the constant of proportionality and how it is obtained. Is it the same in the two situations?

Thank you.
For the original wave,
$$n_p= \frac{c \times \frac{1}{2} \epsilon _0 {E_0}^2}{\hbar \omega}$$
For the compound wave , no. of photons with energy ħω
$$n_{p,\omega}= \frac{c \times \frac{1}{2} \epsilon _0 {E_0}^2}{\hbar \omega}$$
and no. of photons with energy 2ħω
$$n_{p,2 \omega}= \frac{c \times \frac{1}{2} \epsilon _0 ({E_0 /2})^2}{\hbar \times 2 \omega}$$. (per unit cross section area.)
Total incident photons (per second per unit area) for the compound wave
$$n_p'= n_{p,\omega}+n_{p,2 \omega}=\frac{9}{8} \frac{c \times \frac{1}{2} \epsilon _0 {E_0}^2}{\hbar \omega}=\frac{9}{8}n_p$$

 

[kuruman]

That's not what I had in mind. The magnitude of the electric field varies with time, therefore an average needs to be taken over a cycle. The average intensity is proportional to $E_0^2$. Does the average intensity for the two waves have the same proportionality constant?

 

[Terry Bing]

That's not what I had in mind. The magnitude of the electric field varies with time, therefore an average needs to be taken over a cycle. The average intensity is proportional to $E_0^2$. Does the average intensity for the two waves have the same proportionality constant?

It does, doesn't it ? Intensity depends on the amplitude and the wave velocity, which is the same for the two components.
When you say 'the two waves', do you mean the two sinusoidal components of the 2nd wave? This is what I thought you meant. Or do you mean the pure sinusoidal wave vs the compound wave?
Isn't what I wrote down in the previous reply correct? The average intensity is proportional to the amplitude for both the components, with the same proportionality constant, but the energy per photon is different. This is where I made a mistake earlier.

 

[kuruman]

You have to consider that for every cycle of the $\omega$ wave you have two cycles of the $2\omega$ wave. So you should weigh the latter twice in order to find the appropriate fraction.