Animated Solution for Physics - Dual Nature of Matter and Radiation: The electric field of light wave is given as $\mathbf{E}=10^{-3}\cos \left(\frac{2\pi x}{5\times 10^{-7}}-2\pi \times 6\times 10^{14}t\right)\hat{x}{\text{ NC}}^{-1}$. This light falls on a metal plate of work function $2\text{ eV}$. The stopping potential of the photoelectrons is $\left[\text{Given, }E\text{ (in eV)}=\frac{12375}{\lambda \text{ (in A˚)}}\right]$

Imagine a beam of light not just as a stream of particles, but as a beautifully oscillating electromagnetic wave traveling through space. This problem presents us with a fascinating scenario where we must bridge the classical description of light—given by its electric field equation—with its quantum behavior when it strikes a metal surface.

The electric field of the incident light is given by the equation:

At first glance, this might look intimidating. But let's take a breath and break it down. This is simply the standard equation of a traveling wave, which takes the general form:

By comparing our given equation with the standard form, we can extract vital physical properties of the light. The term multiplying the spatial variable $x$ is the wave number, denoted by $k$.

Looking closely, we see that:

We also know from wave mechanics that the wave number $k$ is directly related to the wavelength $\lambda$ by the relation $k=\frac{2\pi }{\lambda }$. Equating these two expressions for $k$, we get:

The $2\pi$ terms elegantly cancel out, revealing the wavelength of our incident light:

To make our upcoming calculations smoother, let's convert this wavelength into Angstroms ($\text{A˚}$). Since $1\text{ m}=10^{10}\text{ A˚}$, we multiply by $10^{10}$ to get:

Calculating Photon Energy

Now that we know the wavelength, we shift our perspective from classical waves to quantum photons. The problem kindly provides a shortcut formula to calculate the energy of these photons directly in electron volts (eV):

Substituting our wavelength of $5000\text{ A˚}$ into this formula, we find the energy of a single incident photon:

When these photons, each carrying $2.475\text{ eV}$ of energy, strike the metal plate, they interact with the electrons inside. The metal has a work function $\varphi =2\text{ eV}$, which is the minimum energy required just to pull an electron out of the metal surface.

According to Einstein's photoelectric equation, the energy of the incident photon is conserved and split into two parts: overcoming the work function, and imparting kinetic energy to the ejected electron.

We are asked to find the stopping potential, $V_0$. The stopping potential is the exact negative voltage required to stop even the fastest, most energetic electrons from reaching the anode. Therefore, the maximum kinetic energy is equal to the work done against this stopping potential:

Substituting this into Einstein's equation, we get:

Stopping Potential

Now, it's just a matter of plugging in the numbers we've discovered. We know $E=2.475\text{ eV}$ and $\varphi =2\text{ eV}$.

Rearranging to solve for the stopping potential term:

The elementary charge '$e$' cancels out on both sides, leaving us with the stopping potential in volts:

Looking at our options, we need to round this to two decimal places. Rounding $0.475\text{ V}$ gives us $0.48\text{ V}$, which perfectly matches option (a).

This problem is a beautiful demonstration of how different domains of physics—wave optics and quantum mechanics—intertwine to describe the reality of light and matter.