Animated Solution for Chemistry - Atomic Structure: A source of monochromatic radiation of wavelength $400\text{ nm}$ provides $1000\text{ J}$ of energy in $10$ seconds. When this radiation falls on the surface of sodium, $x\times 10^{20}$ electrons are ejected per second. Assume that wavelength $400\text{ nm}$ is sufficient for ejection of electron from the surface of sodium metal. The value of $x$ is ...... . (Nearest integer) $(h=6.626\times 10^{-34}\text{ Js})$

Imagine you are standing in a gentle rainstorm, but instead of water, the sky is pouring down pure energy. This is exactly what happens when a powerful light source shines onto a piece of sodium metal. The light carries energy, and as it strikes the surface, it acts like a barrage of tiny billiard balls, knocking electrons completely out of the metal. This phenomenon is the classic photoelectric effect, a cornerstone of quantum mechanics.

In our specific scenario, we are given a macroscopic view of this energy transfer: a total energy of $1000\text{ J}$ is delivered over a span of $10\text{ s}$. We are also given the microscopic signature of the light: its wavelength $\lambda$ is $400\text{ nm}$. Our mission is to bridge these two worlds to find out exactly how many electrons are being ejected every single second.

Microscopic Packets

First, let's figure out the rate at which energy is hitting the metal. We don't want the total energy over ten seconds; we want the energy arriving every single second. This is the definition of power.

By dividing the total energy by the time, we get:

This means that every second, $100\text{ Joules}$ of light energy bombards our sodium surface.

Now, we must shift our perspective to the quantum realm. This $100\text{ J/s}$ isn't a continuous, smooth wave of energy. It arrives in discrete, indivisible packets called photons. To find out how many of these packets are arriving, we first need to determine the energy carried by just one single photon.

We use the famous Planck-Einstein relation:

Let's plug in the fundamental constants. Planck's constant $h$ is $6.626\times 10^{-34}\text{ Js}$, the speed of light $c$ is $3\times 10^8\text{ m/s}$, and our wavelength $\lambda$ is $400\text{ nm}$, which we must convert to standard SI units as $400\times 10^{-9}\text{ m}$.

When we calculate this, we find the energy of a single photon is roughly $4.97\times 10^{-19}\text{ Joules}$. This is an incredibly tiny amount of energy, which tells us there must be a massive number of photons arriving every second to make up the full $100\text{ Joules}$!

Here is the crucial physical assumption: the problem states the wavelength is "sufficient for ejection". This implies that every single photon that hits the surface has enough energy to overcome the metal's work function and eject exactly one electron. We are assuming a 100% quantum efficiency.

Therefore, the number of electrons ejected per second ($n$) is simply the total energy arriving per second divided by the energy of a single photon.

Let's substitute our raw values. Keeping the fraction intact helps us avoid early rounding errors:

Rearranging the terms, the denominator of the bottom fraction flips up to the numerator:

After carefully executing the arithmetic, we find:

The question tells us that the number of electrons ejected per second is formatted as $x\times 10^{20}$, and it asks for the nearest integer value of $x$.

Comparing our calculated result of $2.01\times 10^{20}$ with the given format, it is clear that $x\approx 2.01$. Rounding to the nearest integer, we get our final answer:

By seamlessly connecting the macroscopic power of a light source to the microscopic energy of individual photons, we've successfully counted the invisible quantum rain!