Animated Solution for Physics - Dual Nature of Matter and Radiation: The graph between $1/\lambda$ and stopping potential ($V$) of three metals having work functions ${\varphi }_1,{\varphi }_2$ and ${\varphi }_3$ in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct ? (Here, $\lambda$ is the wavelength of the incident ray).

A Quantum Leap

The photoelectric effect is one of the most beautiful phenomena in modern physics, serving as the bedrock for quantum mechanics. When light of a sufficient frequency shines on a metal surface, it can eject electrons. Einstein brilliantly explained this by proposing that light consists of discrete packets of energy called photons. The energy of each photon is given by $E=\frac{hc}{\lambda }$.

When a photon strikes an electron, it transfers its entire energy. A portion of this energy is used to overcome the binding energy of the electron to the metal, known as the work function $\varphi$. The remaining energy manifests as the maximum kinetic energy of the ejected photoelectron. This relationship is elegantly captured in Einstein's photoelectric equation:

To measure this maximum kinetic energy experimentally, we apply a reverse voltage, known as the stopping potential $V$. When the stopping potential is just enough to halt the fastest electrons, we have $K_{\max }=eV$. Substituting this into our equation yields:

Rearranging this to isolate $V$, we get:

This equation is a masterpiece. It perfectly mirrors the equation of a straight line, $y=mx+c$. Here, our y-axis is the stopping potential $V$, and our x-axis is the inverse of the wavelength, $1/\lambda$.

A Universal Constant

By comparing our rearranged equation with $y=mx+c$, we can immediately identify the slope $m$ of the graph:

Notice something remarkable here: the slope depends only on Planck's constant $h$, the speed of light $c$, and the elementary charge $e$. It is completely independent of the material! This is why the lines for all three metals in the graph are perfectly parallel. Since $\tan \theta =\frac{hc}{e}$, it is directly proportional to $\frac{hc}{e}$ (with a proportionality constant of 1). This confirms that statement (c) is absolutely correct.

Unveiling the Work Function

Now, let's look at the x-intercepts. The x-intercept occurs where the stopping potential $V=0$. This specific point corresponds to the threshold wavelength ${\lambda }_0$, the maximum wavelength that can still cause photoelectric emission. Setting $V=0$ in our equation gives:

This tells us that the work function $\varphi$ is directly proportional to the x-intercept $\frac{1}{{\lambda }_0}$. Let's read the intercepts from the graph for our three metals:

- Metal 1: $\frac{1}{{\lambda }_{01}}=0.001{\text{ nm}}^{-1}$ - Metal 2: $\frac{1}{{\lambda }_{02}}=0.002{\text{ nm}}^{-1}$ - Metal 3: $\frac{1}{{\lambda }_{03}}=0.004{\text{ nm}}^{-1}$

Taking the ratio of their work functions:

This perfectly matches statement (a), making it correct as well.

Who Gets Ejected?

Finally, let's evaluate statement (d). Violet light sits at the high-energy end of the visible spectrum, with a wavelength ${\lambda }_v\approx 400\text{ nm}$. To see where this falls on our graph, we calculate its inverse:

For photoelectric emission to occur, the energy of the incident light must exceed the work function. In terms of our graph, the incident $\frac{1}{\lambda }$ must be greater than the threshold $\frac{1}{{\lambda }_0}$. Let's check each metal:

- Metal 1: $0.0025>0.001$ (Emission occurs!) - Metal 2: $0.0025>0.002$ (Emission occurs!) - Metal 3: $0.0025<0.004$ (No emission!)

Violet light has enough energy to eject electrons from metals 1 and 2, but it falls short for metal 3. Therefore, statement (d) is incorrect. The final correct statements are (a) and (c).