Problem 23 - Master Resource Book in JEE Main Physics

Question

A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become (a) four times the original value (b) twice the original value (c) \(1 / 6 \mathrm{th}\) of the original value (d) unchanged

Step-by-Step Solution

Answer

The maximum kinetic energy remains unchanged (option d).

Step 1: Understand Photoelectric Effect

The photoelectric effect is described by Einstein's equation, \( K_{max} = hf - \Phi \), where \( K_{max} \) is the maximum kinetic energy of photoelectrons, \( h \) is Planck's constant, \( f \) is the frequency of the incident light, and \( \Phi \) is the work function of the metal. The kinetic energy depends only on the light frequency and work function, not on the intensity.

Step 2: Assess the Role of Intensity

While intensity refers to the number of photons striking the surface per unit time, it affects the number of electrons emitted but not their kinetic energy, as kinetic energy is dependent on the energy of individual photons (given by the light's frequency).

Step 3: Evaluate Changes in Conditions

Reducing the intensity to one-fourth means fewer photons are hitting the surface, so fewer electrons might be emitted, but each photon's energy remains the same. Thus, the kinetic energy of any emitted electron does not change.

Step 4: Determine the Correct Option

Since the maximum kinetic energy \( K_{max} \) is unaffected by changes in light intensity, the original kinetic energy remains unchanged. Therefore, (d) "unchanged" is the correct answer.

Key Concepts

Kinetic Energy of Photoelectrons

Kinetic energy in the context of the photoelectric effect indicates how fast the ejected electrons move after absorbing photons. When light hits a metal surface, photons transfer energy to electrons. If this energy surpasses the metal's threshold, or work function, electrons can escape.

The kinetic energy of these electrons depends on two factors: the frequency of the light and the work function of the metal. This energy is described by Einstein's photoelectric equation:

• \( K_{max} = hf - \Phi \)
• \( K_{max} \) is the maximum kinetic energy of the ejected electrons.
• \( h \) is Planck's constant.
• \( f \) is the frequency of the incident light.
• \( \Phi \) is the metal's work function.

Changes in intensity do not affect the kinetic energy of photoelectrons as they depend solely on frequency and work function. Thus, intensity changes alter the number of emitted electrons rather than their velocity.

Einstein's Photoelectric Equation

Einstein's photoelectric equation is a pivotal part of understanding how photoelectrons gain energy when light shines on a metal surface.

This equation, \( K_{max} = hf - \Phi \), forms the basis for quantifying the maximum kinetic energy of photoelectrons:

• The term \( hf \) represents the energy of the incident light. Here, \( h \) is known as Planck's constant, and \( f \) is the frequency of the light.
• \( \Phi \) stands for the work function, which is the minimum energy needed to liberate an electron from the metal's surface.
• Only if \( hf \) surpasses \( \Phi \), will the electron be emitted with excess energy seen as kinetic energy.

This equation elucidates that the frequency of light is intertwined with the available kinetic energy, not the intensity of the light. This is a crucial point in solving for maximum kinetic energy, showing that energy transfer depends heavily on photon frequency.

Intensity and Frequency of Light

Understanding how light's intensity and frequency affect the photoelectric effect is critical.

**Intensity**

• Intensity refers to the amount of energy the light carries per time unit and area. In simpler terms, it's about the number of photons hitting the surface.
• Higher intensity means more photons striking the surface, potentially releasing more electrons.
• However, intensity alone doesn't change the kinetic energy of ejected electrons.

**Frequency**

• The frequency of light is the number of oscillations that light waves complete in a second.
• Higher frequency light carries more energy in each photon, which can increase the excess energy converted into kinetic energy after exceeding the work function barrier.
• Thus, frequency essentially governs the energy imparted to each electron when light hits the metal.

In the given problem, reducing intensity lowers the number of photoelectrons emitted but leaves their kinetic energy constant. Frequency remains the driver of energy per photon, which aligns with Einstein's photoelectric equation.