The de-Broglie wavelength (lambda) associated with a photoelectron varies with the frequency (nu) of

Q: Why do we use kinetic energy instead of momentum in the de-Broglie formula?

Because the photoelectric equation gives us the kinetic energy directly. By converting momentum to kinetic energy using $p = \sqrt{2m(K.E.)}$, we can easily link the two concepts.

Q: What happens if the incident frequency $\nu$ is less than the threshold frequency $\nu_0$?

If $\nu < \nu_0$, the photon doesn't have enough energy to overcome the work function. No electron is ejected, so there is no moving particle and therefore no de-Broglie wave to measure.

Q: Why is the mass $m$ considered constant here?

We are assuming the ejected electron is moving at non-relativistic speeds (much slower than the speed of light). In this regime, its rest mass $m$ remains effectively constant.

Q: Can this formula be applied to protons?

Protons aren't ejected via the photoelectric effect (which specifically involves electrons in a metal lattice). However, the general de-Broglie formula $\lambda = h/p$ applies to protons and all other matter.

Q: Why is the power $1/2$ in the final answer?

The power of $1/2$ comes from the square root in the momentum-kinetic energy relation ($p = \sqrt{2mK.E.}$), which places the kinetic energy term under a square root in the denominator.

Question

The de-Broglie wavelength ($\lambda$) associated with a photoelectron varies with the frequency ($
u$) of the incident radiation as,
[$
u_0$ is threshold frequency]

Options:

(A) $\lambda \propto \frac{1}{(
u - 
u_0)^{1/4}}$

(B) $\lambda \propto \frac{1}{(
u - 
u_0)^{3/2}}$

(C) $\lambda \propto \frac{1}{(
u - 
u_0)}$

(D) $\lambda \propto \frac{1}{(
u - 
u_0)^{1/2}}$

Sigma Percentile · Accepted Answer

The universe is full of beautiful symmetries, and one of the most profound is the dual nature of light and matter. In this problem, we are going to witness a spectacular crossover between two of the greatest ideas in modern physics: Einstein's Photoelectric Effect and de-Broglie's Wave-Particle Duality.

Imagine you are shining a beam of light onto a piece of metal. What happens at the microscopic level? Let's dive in and find out!

### The Photoelectric Effect: Breaking Free

When a photon of light strikes a metal surface, it acts like a tiny packet of energy. If this energy, given by $h
u$, is greater than the metal's work function $h
u_0$, the photon can knock an electron completely out of the metal.

But what happens to the leftover energy? It doesn't just disappear! By the law of conservation of energy, the excess energy becomes the kinetic energy of the ejected electron. This gives us Einstein's famous photoelectric equation:

$$K.E. = h
u - h
u_0$$

Here, $
u$ is the frequency of the incident light, and $
u_0$ is the threshold frequency of the metal. This equation is our first key to unlocking the problem.

### The de-Broglie Wavelength: Riding the Wave

Now, let's shift our perspective. Once the electron is flying through space with that kinetic energy, it doesn't just act like a tiny billiard ball. According to Louis de-Broglie, every moving particle has a wave associated with it!

The wavelength of this matter wave is inversely proportional to the particle's momentum. We can write the de-Broglie wavelength $\lambda$ in terms of kinetic energy as:

$$\lambda = \frac{h}{\sqrt{2m(K.E.)}}$$

This is our second master equation. It beautifully connects the mechanical property of kinetic energy to the wave property of wavelength.

### The Grand Synthesis

Now comes the magic. We have the kinetic energy from the photoelectric effect, and we need the wavelength from de-Broglie's hypothesis. Let's substitute the first equation into the second:

$$\lambda = \frac{h}{\sqrt{2m(h
u - h
u_0)}}$$

Let's clean this up by factoring out Planck's constant $h$ in the denominator:

$$\lambda = \frac{h}{\sqrt{2mh(
u - 
u_0)}}$$

Look at the structure of this equation. The terms $h$, $m$, and $2$ are all constants. They don't change when we vary the frequency of the incident light. Therefore, we can strip away the constants to find the direct proportionality:

$$\lambda \propto \frac{1}{\sqrt{
u - 
u_0}}$$

Since a square root is mathematically equivalent to a power of $1/2$, we can rewrite this as:

$$\lambda \propto \frac{1}{(
u - 
u_0)^{1/2}}$$

**This perfectly matches option (d)!**

By simply combining two fundamental equations, we've discovered exactly how the matter wave of an ejected electron stretches and compresses as we change the color of the light shining on the metal. It's a brilliant reminder of how interconnected the laws of quantum mechanics truly are!

Animated Solution for Chemistry - Atomic Structure: The de-Broglie wavelength ( $λ$ ) associated with a photoelectron varies with the frequency ( $ν$ ) of the incident radiation as, [ $ν_{0}$ is threshold frequency]

Select Answer:

Visualized Solution

Photoelectric Effect Setup

Einstein's Photoelectric Equation

de-Broglie Wavelength

Substituting Kinetic Energy

Finding the Proportionality

Final Relation

What if we plot a graph?

The Sigma Insight: Wave Particle Duality

The Photoelectric Effect