Animated Solution for Chemistry - Atomic Structure: When light of wavelength $248\text{ nm}$ falls on a metal of threshold energy $3.0\text{ eV}$, the de-Broglie wavelength of emitted electrons is ...... $\text{A˚}$. [Round off to the nearest integer] [Use: $\sqrt{3}=1.73$, $h=6.63\times 10^{-34}\text{ Js}$, $m_e=9.1\times 10^{-31}\text{ kg}$, $c=3.0\times 10^8{\text{ ms}}^{-1}$, $1\text{ eV}=1.6\times 10^{-19}\text{ J}$]

Imagine you are observing a microscopic game of billiards. A photon of light, acting like a tiny energy packet, crashes into a metal surface. If this photon packs enough punch, it can knock an electron completely out of the metal. This beautiful phenomenon is the Photoelectric Effect, and it perfectly bridges the gap between light acting as a wave and as a particle.

In our problem, we are shining light with a wavelength of $248\text{ nm}$ onto a metal that has a threshold energy (or work function, $\varphi$) of $3.0\text{ eV}$. Our ultimate mission is to find the de-Broglie wavelength of the electron that gets ejected.

To understand what happens to the electron, we first need to know exactly how much energy our incoming photon is carrying. The standard formula is $E=\frac{hc}{\lambda }$. However, plugging in Planck's constant and the speed of light in standard SI units is a recipe for calculation errors during a high-pressure exam.

Instead, we use a brilliant shortcut. If you express the wavelength in Angstroms ($\text{A˚}$), the energy in electron-volts ($\text{eV}$) is simply:

First, let's convert our wavelength: $248\text{ nm}=2480\text{ A˚}$.

Now, substitute this into our shortcut:

Our incident photon arrives with exactly $5\text{ eV}$ of energy.

Now we bring in Einstein's Photoelectric Equation. The total energy of the photon is spent doing two things: paying the "toll" to escape the metal (the work function), and giving the electron its kinetic energy.

We know the photon brings $5\text{ eV}$, and the metal demands $3.0\text{ eV}$.

The ejected electron flies away with a maximum kinetic energy of $2.0\text{ eV}$.

Here is where the magic happens. Just as light can act as a particle, our ejected electron (a particle) acts as a wave! Its wavelength is given by the de-Broglie relation:

Once again, calculating this with raw SI units is tedious. For an electron specifically, we have another powerful shortcut. If the kinetic energy is known in $\text{eV}$ (which is numerically equal to the accelerating potential $V$ in volts), the wavelength in Angstroms is:

Since our electron has a kinetic energy of $2.0\text{ eV}$, we can set $V=2$. Let's plug this into our shortcut:

We know that $8^2=64$ and $9^2=81$. Therefore, $\sqrt{75}$ must be somewhere in between, roughly $8.66\text{ A˚}$.

The question explicitly asks us to round off to the nearest integer. Since $8.66$ is closer to $9$ than it is to $8$, we round up.

Always remember to read the rounding instructions carefully. A small oversight there can cost you a perfectly solved question!