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The intensity of the electric field of an electromagnetic wave varies according to the law:$$E\left(t\right)=E_{0}\left(1+\cos\varOmega t\right)\cdotp \cos\omega t,\varOmega=2\cdotp10^{15}s^{-1},\omega=2\cdotp10^{16}s^{-1}.$$
Find the energy of the electrons, expelled by this wave from the hydrogen atoms, whose ionisation energy is $W=13,6(eV)$.

I tried to express this signal(using amplitude modulation) as the sum of three harmonic components:$$E\left(t\right)=E_{0}\cdotp\left(1+\cos\varOmega t\right)\cdotp \cos\omega t=E_{0}\cos\omega t+\frac{E_{0}}{2}\cos\left(\omega-\Omega\right)t+\frac{E_{0}}{2}\cos\left(\omega+\varOmega\right)t .$$ How to proceed from here? I don't know how to use this to find what the problem asks for.

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closed as off-topic by Emilio Pisanty, John Rennie, Kyle Kanos, Jon Custer, tpg2114 Jul 10 at 15:09

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    $\begingroup$ Welcome to Physics! Please note that this site is not a place to obtain solutions to worked problems. Please see this Meta post on asking homework-like questions and this Meta post for "check my work problems". $\endgroup$ – Kyle Kanos Jul 9 at 12:59
  • $\begingroup$ @KyleKanos this is by no means a homework-like question in my opinion. I didn't even tag this as "homework and exercises", this was done by someone else. The question posted by me was asked in 2017 in Moldova on the first team selection test for the International Physics Olympiad. I presented my attempt at a solution and from that point I am stuck. $\endgroup$ – ChemistryGeek Jul 9 at 15:25
  • $\begingroup$ Considering everything after your second sentence in your above comment fits the exact definition of "homework-like" on this site, you may need to update your opinion. We don't help people who get stuck solving a problem, as explained in my two links. $\endgroup$ – Kyle Kanos Jul 9 at 17:45
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Once you have expressed the electric field in this form, you easily see that the angular frequencies of the photons carried by the field are $\omega$, $\omega$ - $\Omega$ and $\omega$ + $\Omega$. The energy of the photons can be calculated from $ E = \hbar \omega$.

Now if we ignore multi-photon interactions, the energy of the electrons expelled would correspond to the energy difference between the energy of the photons and the hydrogen atom electronic levels. From a rough calculation, one can also see that only the photons with angular frequency $\omega$ + $\Omega$ are energetic enough to knock electrons out of the ground state of hydrogen. However, if excited hydrogen atoms are also present, one can expect electrons expelled with a range of different discrete energy values.

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