Animated Solution for Physics - Dual Nature of Matter and Radiation: A beam of light has three wavelengths $4144\text{ A˚}$, $4972\text{ A˚}$ and $6216\text{ A˚}$ with a total intensity of $3.6\times 10^{-3}{\text{ W m}}^{-2}$ equally distributed amongst the three wavelengths. The beam falls normally on an area $1.0{\text{ cm}}^2$ of a clean metallic surface of work function $2.3\text{ eV}$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds.

The photoelectric effect is one of the most fascinating phenomena in modern physics. It’s the very experiment that proved light behaves as a particle—a photon. Imagine the surface of a metal as a highly exclusive club, and the electrons are the guests inside. The work function of the metal is the entry fee (or rather, the exit fee) they need to pay to leave. The incident photons are like friends throwing money over the wall. If a single photon doesn't have enough energy to cover the work function, the electron can't leave. It doesn't matter how many low-energy photons hit the surface; they can't pool their money together. It's a strict one-to-one interaction!

In this beautiful problem from JEE Advanced 1989, we are given a beam of light consisting of three different wavelengths, and we need to find out exactly how many electrons manage to escape the metal surface in a span of two seconds. Let's break down the physics and the math step-by-step.

We are told that the incident light beam is a mixture of three wavelengths: 1. ${\lambda }_1=4144\text{ A˚}$ 2. ${\lambda }_2=4972\text{ A˚}$ 3. ${\lambda }_3=6216\text{ A˚}$

The total intensity of this combined beam is $I_{total}=3.6\times 10^{-3}{\text{ W/m}}^2$. A crucial piece of information here is that this intensity is equally distributed among the three wavelengths.

This means each individual wavelength carries exactly one-third of the total intensity. Let's calculate the intensity $I$ for each specific wavelength:

The beam falls normally on a metallic surface with an area $A=1.0{\text{ cm}}^2$. Before we proceed, we must convert this area into standard SI units (square meters) to avoid any silly mistakes later on.

Now, we can find the power $P$ (energy per second) delivered by each wavelength to the given area. Power is simply intensity multiplied by area:

So, every second, each of the three wavelengths delivers $1.2\times 10^{-7}\text{ Joules}$ of energy to the metal surface.

The Bouncer at the Door

Now comes the most critical conceptual step. Just because a wavelength is delivering energy to the surface doesn't mean it will eject electrons. We must check if the individual photons of each wavelength have enough energy to overcome the metal's work function, $W=2.3\text{ eV}$.

The energy of a single photon is given by the Planck-Einstein relation:

For quick calculations in electron-volts (eV) when the wavelength is in Angstroms ($\text{A˚}$), we use the highly useful approximation $hc\approx 12400\text{ eV A˚}$. However, for more precise calculations (as often required in older JEE papers), using $hc=12375\text{ eV A˚}$ yields more accurate results. Let's calculate the energy for each wavelength.

For the first wavelength (${\lambda }_1=4144\text{ A˚}$):

For the second wavelength (${\lambda }_2=4972\text{ A˚}$):

For the third wavelength (${\lambda }_3=6216\text{ A˚}$):

Who Gets Ejected?

Now, we compare these photon energies with the work function $W=2.3\text{ eV}$.

- $E_1(2.99\text{ eV})>2.3\text{ eV}$: These photons have more than enough energy. They will successfully eject electrons. - $E_2(2.49\text{ eV})>2.3\text{ eV}$: These photons also cross the threshold. They will eject electrons too. - $E_3(1.99\text{ eV})<2.3\text{ eV}$: Here is the catch! These photons are too weak. No matter how many of them hit the surface, they cannot eject a single electron.

This is a classic trap in JEE problems. We must completely ignore the third wavelength for the rest of our calculations regarding photoelectron emission. It contributes to the heating of the metal, but not to the photoelectric current.

We know that the first two wavelengths are capable of ejecting electrons. The problem states that "each energetically capable photon ejects one electron." This means the quantum efficiency is $100\%$. Therefore, the number of electrons emitted per second is exactly equal to the number of capable photons hitting the surface per second.

Let $n_1$ be the number of photons of the first wavelength hitting the surface per second. We can find this by dividing the total power of that wavelength by the energy of a single photon.

CRITICAL WARNING: The power is in Joules per second (Watts), so the photon energy must also be converted from eV to Joules! ($1\text{ eV}=1.6\times 10^{-19}\text{ J}$)

Similarly, let $n_2$ be the number of photons of the second wavelength hitting the surface per second:

We now have the emission rates caused by both capable wavelengths. The total number of electrons emitted per second is simply the sum of these two rates:

The question asks for the total number of photoelectrons liberated in two seconds. We just need to multiply our rate by the time interval:

And there we have it! By carefully analyzing the intensity distribution, filtering out the incapable photons, and maintaining strict unit consistency, we arrive at the final answer. This problem beautifully illustrates the quantum nature of light and the threshold condition of the photoelectric effect.