Problem 8 - Master Resource Book in JEE Main Physics

Question

Maximum velocity of photoelectron emitted is \(4.8 \mathrm{~ms}^{-1}\). The \(\frac{e}{m}\) ratio of electron is \(1.76 \times 10^{11} \mathrm{Ckg}^{-1}\), then stopping potential is given by (a) \(5 \times 10^{-10} \mathrm{JC}^{-1}\) (b) \(3 \times 10^{-7} \mathrm{JC}^{-1}\) (c) \(7 \times 10^{11} \mathrm{JC}^{-1}\) (d) \(2.5 \times 10^{-2} \mathrm{JC}^{-1}\)

Step-by-Step Solution

Answer

The stopping potential is \(2.5 \times 10^{-2} \mathrm{JC}^{-1}\) (Option d).

Step 1: Understanding the Problem

We are given the maximum velocity of a photoelectron and the charge-to-mass ratio of an electron. We need to find the stopping potential, which is related to the kinetic energy of the photoelectron.

Step 2: Finding Kinetic Energy of the Photoelectron

The kinetic energy (KE) of the photoelectron can be calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron, and \( v \) is the velocity of the electron (\( 4.8 \text{ ms}^{-1} \)). The ratio \( \frac{e}{m} \) relates charge \( e \) and mass \( m \) as follows: \( m = \frac{e}{\frac{e}{m}} \).

Step 3: Substituting the Values

Substitute to find \( m \) from \( \frac{e}{m} = 1.76 \times 10^{11} \text{ Ckg}^{-1} \): \( m = \frac{e}{1.76 \times 10^{11}} \). Use this \( m \) in the KE formula: \( KE = \frac{1}{2} \cdot \frac{e}{1.76 \times 10^{11}} \cdot (4.8)^2 \).

Step 4: Calculating Stopping Potential

The stopping potential \( V_s \) is related to the kinetic energy \( KE \) by \( KE = eV_s \). Solve for \( V_s \) by rearranging the equation: \( V_s = \frac{KE}{e} \). Each \( e \) terms will cancel out, giving \( V_s = \frac{(4.8)^2}{2 \cdot 1.76 \times 10^{11}} \).

Step 5: Final Calculation

Compute the numerical value: \( V_s = \frac{4.8^2}{2 \times 1.76 \times 10^{11}} \approx 2.5 \times 10^{-2} \). Check this value against the provided options.

Key Concepts

Stopping Potential

The stopping potential is a crucial concept in the photoelectric effect. In simple terms, it's the minimum potential difference needed to stop the most energetic photoelectrons from reaching the collector electrode. These electrons are emitted from a material when it's exposed to light, like ultraviolet or visible light.

For the photoelectric effect, light energy hits a material, releasing electrons. These electrons have kinetic energy, which is linked to the stopping potential. The formula relating them is \[ KE = eV_s \] where \( KE \) is the kinetic energy of the emitted photoelectron, \( e \) is the charge of an electron, and \( V_s \) is the stopping potential. The idea is that the stopping potential decelerates the photoelectrons to a halt, basically using an electric field to counter their kinetic energy.

Understanding stopping potential helps in exploring how electromagnetic radiation releases electrons from different materials.

Kinetic Energy of Electrons

Kinetic energy is the energy an object possesses due to its motion. In the context of the photoelectric effect, it's the energy that photoelectrons gain as they are emitted from a material surface after being struck by light. The formula for kinetic energy is straightforward:

\[ KE = \frac{1}{2}mv^2 \] Here, \( m \) is the mass of the electron and \( v \) is the velocity. In photoelectric experiments, measuring this velocity allows scientists to determine the kinetic energy of electrons.

What's interesting is that this kinetic energy is directly related to the light frequency that strikes the material. The excess energy, after overcoming the work function (minimum energy needed to release an electron), is converted into kinetic energy for the photoelectrons. This is crucial for understanding how light can transfer energy to electrons. Once you know the kinetic energy, you can easily compute the stopping potential, creating a bridge between light energy input and electronic energy output.

Charge-to-Mass Ratio

The charge-to-mass ratio, often denoted as \( \frac{e}{m} \), is an essential characteristic of electrons. It determines how an electron responds to electric and magnetic fields. In the context of our exercise, understanding this ratio is critical as it links the physical properties to the forces acting upon the electrons.

Knowing \( \frac{e}{m} = 1.76 \times 10^{11} \text{ Ckg}^{-1} \), for instance, helps you calculate other electron properties such as their kinetic energy when maximum velocity is known. This ratio simplifies complex calculations by letting you express mass \( m \) as \( \frac{e}{\frac{e}{m}} \).

This concept is not only useful in theoretical physics problems like the one mentioned but also practical scenarios like designing electron-beam devices. When physicists discovered the charge-to-mass ratio of the electron, it was a significant leap in understanding atomic and particle physics, as it is fundamental to mathematics underpinning electronic properties.