Animated Solution for Physics - Dual Nature of Matter and Radiation: The magnetic field associated with a light wave is given at the origin, by $B=B_0[\sin (3.14\times 10^7)ct+\sin (6.28\times 10^7)ct]$. If this light falls on a silver plate having a work function of $4.7\text{ eV}$, what will be the maximum kinetic energy of the photoelectrons? (Take, $c=3\times 10^8{\text{ ms}}^{-1}$ and $h=6.6\times 10^{-34}\text{ J-s}$)

A Beautiful Bridge

Imagine you are standing at the crossroads of classical electromagnetism and quantum mechanics. On one side, you have James Clerk Maxwell describing light as a continuous electromagnetic wave. On the other side, you have Albert Einstein describing light as a stream of discrete energy packets called photons. This problem is a beautiful bridge between these two worlds.

We are given the magnetic field equation of a light wave:

At first glance, this looks like pure classical physics. But when this light hits a silver plate, it triggers the photoelectric effect—a purely quantum phenomenon. Our mission is to find the maximum kinetic energy of the ejected electrons.

Let's break down the magnetic field equation. Notice that it contains two sine terms. This means the incident light is not monochromatic; it is a superposition of two waves with different frequencies.

In the quantum picture, this implies that the light beam consists of two different types of photons, each carrying a different amount of energy.

When these photons bombard the silver plate, they will eject electrons. However, the question asks for the maximum kinetic energy of the photoelectrons. According to Einstein's photoelectric theory, the kinetic energy of an ejected electron depends solely on the energy of the individual photon that struck it. Therefore, to find the maximum kinetic energy, we must identify the most energetic photons in the beam.

Energy is directly proportional to frequency ($E = h u$). Thus, our first task is to find the higher frequency component from the wave equation.

The standard wave equation is often written in the form $\sin (kx-\omega t)$. At the origin ($x=0$), this becomes $\sin (-\omega t)$ or simply $\sin (\omega t)$ if we ignore the phase sign.

In our given equation, the terms are $\sin (\text{constant}\cdot ct)$. Since the speed of light $c=\omega /k$, we can deduce that the constant multiplying $t$ is indeed the angular frequency $\omega$.

Let's extract the two angular frequencies:

It is crystal clear that ${\omega }_2$ is the higher angular frequency. This is the one that will produce the most energetic photoelectrons.

Energy of a Photon

Now, let's calculate the actual frequency $ u$ corresponding to ${\omega }_2$. We know that $\omega = 2\pi u$, so:

With the frequency in hand, we can find the energy of these high-octane photons using Planck's equation:

Substituting the given value of Planck's constant ($h=6.6\times 10^{-34}\text{ J-s}$):

Since the work function of the silver plate is given in electron-volts (eV), we must convert our photon energy into the same units to make them compatible. We do this by dividing by the elementary charge ($1.6\times 10^{-19}\text{ C}$):

We have finally reached the climax of the problem. We know the energy of the incoming photons ($12.375\text{ eV}$), and we know the "toll" required to escape the silver plate, which is its work function ($\varphi =4.7\text{ eV}$).

Einstein's photoelectric equation elegantly states that the maximum kinetic energy of the ejected electron is whatever energy is left over after paying the toll:

Substituting our values:

Looking at our options, the closest value is $7.72\text{ eV}$.

A quick note on precision: If we had used the more precise value of Planck's constant ($h=6.626\times 10^{-34}\text{ J-s}$), our photon energy would have been exactly $12.42\text{ eV}$, leading to a kinetic energy of exactly $7.72\text{ eV}$. In competitive exams, it is common to use the simplified values provided in the question text for speed, but always choose the closest matching option if a slight rounding discrepancy occurs.