Problem 13 - Master Resource Book in JEE Main Physics

Question

Two identical photo cathodes receive light of frequencies \(f_{1}\) and \(f_{2} .\) If the velocities of the photo electrons (of mass \(m\) ) coming out are respectively, \(v_{1}\) and \(v_{2}\), then (a) \(v_{1}-v_{2}=\left[\frac{2 h}{m}\left(f_{1}-f_{2}\right)\right]^{1 / 2}\) (b) \(v_{1}^{2}-v_{2}^{2}=\frac{2 h}{m}\left(f_{1}-f_{2}\right)\) (c) \(v_{1}+v_{2}=\left[\frac{2 h}{m}\left(f_{1}+f_{2}\right)\right]^{1 / 2}\) (d) \(v_{1}^{2}+v_{2}^{2}=\frac{2 h}{m}\left(f_{1}+f_{2}\right)\)

Step-by-Step Solution

Answer

The correct answer is option (b): \( v_{1}^{2}-v_{2}^{2}=\frac{2 h}{m}\left(f_{1}-f_{2}\right) \).

Step 1: Understand the Photoelectric Equation

The photoelectric effect is described by the equation \( K.E = h f - \phi \), where \( K.E \) is the kinetic energy of the ejected photoelectron, \( h \) is Planck's constant, \( f \) is the frequency of the incident light, and \( \phi \) is the work function of the material.

Step 2: Relate Kinetic Energy to Velocity

The kinetic energy \( K.E \) of an electron can also be expressed using its mass \( m \) and velocity \( v \) as \( K.E = \frac{1}{2} m v^2 \). Thus, the velocity can be found by \( v = \sqrt{\frac{2 K.E}{m}} \).

Step 3: Apply the Photoelectric Equation for Both Frequencies

For frequency \( f_1 \), the kinetic energy is \( K.E_1 = h f_1 - \phi \) and for \( f_2 \), it is \( K.E_2 = h f_2 - \phi \). These correspond to velocities \( v_1 = \sqrt{\frac{2 (h f_1 - \phi)}{m}} \) and \( v_2 = \sqrt{\frac{2 (h f_2 - \phi)}{m}} \).

Step 4: Analyze the Expression for Velocity Differences

Consider the difference in velocities squared: \( v_1^2 - v_2^2 = \frac{2 (h f_1 - \phi)}{m} - \frac{2 (h f_2 - \phi)}{m} \), simplifying to \( v_1^2 - v_2^2 = \frac{2 h}{m}(f_1 - f_2) \), which matches option (b).

Step 5: Review Other Possibilities

No other options match directly in terms of physical principles using the basic equations derived: (a), (c), and (d) do not correspond mathematically directly from the derivations.

Key Concepts

Kinetic Energy

In the context of the photoelectric effect, kinetic energy is the energy possessed by the electrons that are ejected from a material's surface when it is exposed to light. When light (photons) of a certain frequency strikes a material, it transfers energy to the electrons. The energy transfer is only successful if the photon's energy surpasses a certain threshold called the work function of the material.

Kinetic energy (\(K.E\)) can be expressed using the equation:

• \(K.E = h f - \phi\)

where \(h\) is Planck's constant, \(f\) is the frequency of the light, and \(\phi\) is the work function. The work function is the minimum energy required to liberate an electron from the material. The remaining energy is converted to the kinetic energy of the electron, enabling it to move freely.

Alternatively, kinetic energy can also be related to the velocity \(v\) of the electron and its mass \(m\) using the equation:

• \(K.E = \frac{1}{2} m v^2\)

This relationship allows us to calculate the electron's velocity, providing insights into the energy dynamics in the photoelectric effect. The velocity can be determined by rearranging the kinetic energy formula to solve for \(v\), which is critical for analyzing electron behavior under different frequencies of light.

Planck's Constant

Planck's constant (\(h\)) is a fundamental constant in quantum mechanics, pivotal to various physics equations, including those governing the photoelectric effect. It represents the proportionality factor between the energy of a photon and the frequency of its associated electromagnetic wave.

• \(h = 6.62607015 \times 10^{-34} \text{ Js}\)

Planck's constant is essential when calculating the energy of photons, which is given by the equation:

• Energy \(E = h f\)

In the context of the photoelectric effect, the energy of incoming photons determines if electrons overcome the work function of a material and are ejected. By multiplying Planck's constant by the frequency, we obtain the energy that the photons carry. This value must be sufficient to both liberate the electrons and impart additional kinetic energy to them, allowing them to be detected and measured once ejected.

Understanding \(h\)'s role highlights the quantum nature of light and energy transfer, offering insight into the discrete interactions at an atomic and subatomic level.

Photon Frequency

Photon frequency (\(f\)) is a measure of how many wave cycles of a photon pass a particular point in one second. It is a critical parameter in determining the energy of a photon, which can be calculated using Planck's constant:

• \(E = h f\)

Frequency is directly proportional to the energy of a photon. Higher frequency light will have more energetic photons, which are more likely to eject electrons from a material if they surpass the work function. In the photoelectric effect, varying the frequency of the light affects the kinetic energy of the ejected electrons since higher frequencies provide more energy beyond the work function.

Two different frequencies, \(f_1\) and \(f_2\), will impart different amounts of energy to the electrons. This variance results in differing kinetic energies and velocities for the ejected electrons. Investigating these frequency changes allows us to explore the relationship between photon frequency and electron behavior in photoelectric interactions, an essential concept for applications in solar cells, imaging devices, and more.