Animated Solution for Chemistry - Atomic Structure: A metal surface is exposed to $500\text{ nm}$ radiation. The threshold frequency of the metal for photoelectric current is $4.3\times 10^{14}\text{ Hz}$. The velocity of ejected electron is ......... $\times 10^5{\text{ ms}}^{-1}$. (Nearest integer) [Use $h=6.63\times 10^{-34}\text{ Js}$, $m_e=9.0\times 10^{-31}\text{ kg}$]

Unraveling the Photoelectric Effect

Imagine you are standing on the edge of a microscopic world. A pristine metal surface lies before you, calm and undisturbed. Suddenly, a beam of light—a stream of energetic photons—strikes the surface. The electrons inside the metal, which were previously bound to their atoms, absorb this incoming energy. If the energy is sufficient, they break free and shoot out into the void. This beautiful, almost magical phenomenon is what we call the Photoelectric Effect.

To understand exactly how fast these electrons are moving when they escape, we must turn to the genius of Albert Einstein. His photoelectric equation is a masterpiece of energy conservation.

Einstein proposed that the total energy of an incoming photon ($E$) is split into two distinct parts.

First, a portion of the energy is used simply to tear the electron away from the attractive forces of the metal. This minimum required energy is known as the Work Function ($W$).

Whatever energy is left over is entirely converted into the kinetic energy ($KE$) of the escaping electron. Mathematically, this is expressed as:

Or, expanding the terms:

Let's tackle this problem step by step. First, we need to find out exactly how much energy our incoming photons are carrying. We are given the wavelength of the incident radiation, $\lambda =500\text{ nm}$. Using the Planck-Einstein relation, we calculate the total energy:

After carefully crunching the numbers, we find that the incident energy is $3.978\times 10^{-19}\text{ J}$.

Next, we must determine the toll the metal takes—the work function. We are given the threshold frequency, $ u_0 = 4.3 \times 10^{14} \text{ Hz}$. The work function is simply Planck's constant multiplied by this threshold frequency:

This gives us a work function of $2.8509\times 10^{-19}\text{ J}$.

Calculating Velocity

Now that we know how much energy came in and how much was spent escaping, we can find the leftover kinetic energy by simple subtraction:

This kinetic energy is what drives the electron forward. We know the classical formula for kinetic energy is $\frac{1}{2}m{v}^2$. By equating our calculated energy to this formula, we can solve for the velocity ($v$):

Rearranging the equation to isolate $v^2$, we get:

To make taking the square root easier, let's rewrite this as $25.04\times 10^{10}$. Taking the square root of both sides, we find:

And there we have it! The ejected electron zooms away at a staggering speed of $5\times 10^5\text{ m/s}$. The nearest integer value for our answer is 5.