LEVELJEE Main
Visualized Solution
The Sigma Insight: Photoelectric Effect
The photoelectric effect is one of the most beautiful phenomena in modern physics. It bridges the gap between the classical wave theory of light and the quantum mechanical particle theory. In this problem, we are going to dive deep into the mechanics of this effect, calculating not just the energies involved, but the actual number of particles interacting!
Imagine you are standing in a lab, shining a beam of light onto a pristine platinum surface. We know the intensity of the light, the area it covers, and the energy of each individual "packet" of light—the photon. Our mission is to find out exactly how many electrons are knocked out of the metal every second, and how fast they are moving.
Analyzing the Setup
Let's start by understanding the energy of our incoming light. We are given that each photon carries an energy of .
However, our light intensity is given in , which is a standard SI unit (Joules per second per square meter). To make our calculations seamless, we must convert the photon energy from electron-volts into Joules.
We know the conversion factor: .
This is the energy of a single photon. Now, let's look at the macroscopic picture. How much total energy is hitting our specific piece of platinum every second?
The total power incident on the surface is simply the intensity multiplied by the area .
Counting the Photons
We know the total energy arriving every second, and we know the energy of each individual photon. By dividing the total energy by the energy of a single photon, we can count the exact number of photons striking the surface per second!
That is a massive number of photons! But here is the catch: not every photon is successful in ejecting an electron. The problem states that the efficiency is only .
This means out of all the photons hitting the metal, only a tiny fraction actually knock an electron loose. Let's calculate the number of emitted photoelectrons .
The Master Equation
Now that we know how many electrons are escaping, let's figure out how fast they are moving. For this, we turn to Albert Einstein's famous photoelectric equation.
The maximum kinetic energy of an ejected electron is equal to the energy of the incoming photon minus the work function of the metal (the minimum energy required to break the electron free).
We are given the work function , and we already know the incident photon energy .
This is the energy of the fastest, most energetic electrons that were right at the surface of the metal.
Final Calculation
But what about the minimum kinetic energy?
Imagine an electron that is buried deep within the platinum lattice. When it absorbs a photon, it starts moving towards the surface. However, on its way out, it collides with other atoms and loses some of its energy.
If it loses just enough energy so that it barely escapes the surface, its remaining kinetic energy will be zero. Therefore, the minimum kinetic energy of an emitted photoelectron is always zero.
Final Answer: The number of photoelectrons emitted per second is , with kinetic energies ranging from to .
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