Problem 2 - Master Resource Book in JEE Main Physics

Question

In a photoelectric experiment for 4000 \AA incident radiation. The potential difference to stop the ejection is \(2 \mathrm{~V}\). If the incident light is changed to \(3000 \AA\), then the potential required to stop the ejection of electrons will be (a) greater than \(2 \mathrm{~V}\) (b) less than \(2 \mathrm{~V}\) (c) \(\infty\) (d) zero

Step-by-Step Solution

Answer

(a) Greater than \(2 \mathrm{~V}\).

Step 1: Understanding the Photoelectric Experiment

In a photoelectric experiment, electrons are ejected from a metal surface when light of sufficient frequency (or energy) strikes it. The stopping potential is the minimum energy (potential difference) required to prevent ejected electrons from reaching the other side of the setup. This potential corresponds to the kinetic energy of the ejected electrons.

Step 2: Calculating Energy of Light in First Scenario

Convert the wavelength of the incident light to energy using the formula: \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3.00 \times 10^{8} \text{ m/s} \), and \( \lambda = 4000 \text{ Å} = 4000 \times 10^{-10} \text{ m} \). The energy is calculated in Joules, then converted to electron volts (eV).

Step 3: Calculating Energy of Light in Second Scenario

Similarly, for light of wavelength \( 3000 \text{ Å} = 3000 \times 10^{-10} \text{ m} \), calculate the energy \( E' \) using \( E' = \frac{hc}{\lambda} \). This energy is also converted to electron volts (eV).

Step 4: Comparing Stopping Potential in Both Scenarios

The stopping potential is directly related to the energy of the incident photons. Since \( 3000 \text{ Å} \) light has a shorter wavelength than \( 4000 \text{ Å} \), it means it has more energy. Therefore, the stopping potential required will be greater for light of \( 3000 \text{ Å} \) compared to \( 4000 \text{ Å} \).

Step 5: Conclusion

Based on the energy comparison, more energy means higher stopping potential. Since \( 3000 \text{ Å} \) light has more energy than \( 4000 \text{ Å} \), the stopping potential must be greater than \( 2 \text{ V} \). Thus, option (a) is correct: The potential required is greater than \( 2 \text{ V} \).

Key Concepts

Stopping Potential

The concept of stopping potential is central to understanding the photoelectric effect. When light shines on a metal surface, it can cause electrons to be ejected. However, not all ejected electrons will have enough energy to reach the other side of the apparatus after leaving the surface. The stopping potential is the minimum voltage needed to stop these electrons from completing their journey. In simpler terms, it negates the kinetic energy that the electrons gained from the incident photons.

In the photoelectric experiment scenario, when the stopping potential (given in volts) is applied across the setup, it impedes the movement of electrons between the plates, preventing them from generating a measurable current. Hence, if we want to know if the electrons have enough energy to surpass this potential, we calculate it based on the characteristics of the light that hits the surface.

Incident Photon Energy

The energy of the incident photon plays a crucial role in the photoelectric effect. The energy of a photon is directly linked to its frequency via the equation:
\( E = hf \), where \( h \) is Planck's constant, and \( f \) is the frequency of the photon. However, in many photoelectric experiments, it's more convenient to use the relationship between energy and wavelength:
\( E = \frac{hc}{\lambda} \), where \( \lambda \) is the wavelength.

This formula helps us understand that shorter wavelengths correspond to higher energy photons. So, when a photon with sufficient energy strikes the metal surface, it transfers its energy to an electron, potentially freeing it. Reflecting on the provided experiment, changing the wavelength from 4000 Å to 3000 Å increases the photon's energy, which implies that electrons will be ejected with greater kinetic energy, needing a higher stopping potential to be halted.

Wavelength and Frequency Relationship

Understanding the relationship between wavelength and frequency is crucial for grasping the behavior of electromagnetic waves. Wavelength \( \lambda \) is the distance between successive crests of a wave, whereas frequency \( f \) is how often the wave crests pass a given point per second. These two quantities are inversely related and are connected through the speed of light \( c \):
\( c = \lambda f \).

Thus, as the wavelength decreases, the frequency increases and vice versa. In context to the photoelectric effect, a shorter wavelength means a higher frequency, which leads to higher energy for the photons (since \( E = hf \)).

• This relationship explains why light at 3000 Å (shorter wavelength) has more energy than light at 4000 Å (longer wavelength).
• Ultimately, more energetic photons imply a need for a greater stopping potential to counter the increased kinetic energy of ejected electrons.

These foundational concepts highlight the robustness of the photoelectric effect as it relates to fundamental properties of light and energy transfer.