Problem 3 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

If \(p\) is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength \(\lambda\), then for \(1.5 p\) momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function): [Main April 8, 2019 (II)] (a) \(\frac{3}{4} \lambda\) (b) \(\frac{1}{2} \lambda\) (c) \(\frac{2}{3} \lambda\) (d) \(\frac{4}{9} \lambda\)

Step-by-Step Solution

Answer

The wavelength should be \( \frac{4}{9} \lambda \).

Step 1: Understanding the Relationship

The energy of the photon is given by the equation \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant and \( c \) is the speed of light. The kinetic energy \( KE \) of the ejected electron is related to this energy minus the work function \( \phi \), but since we assume that \( KE \) is high compared to \( \phi \), we can approximate \( KE \approx \frac{hc}{\lambda} \).

Step 2: Express Momentum in Terms of Kinetic Energy

The momentum \( p \) of the electron is given by the relation \( p = \sqrt{2m_e KE} \), where \( m_e \) is the mass of the electron. Substituting \( KE = \frac{hc}{\lambda} \), the relation becomes \( p = \sqrt{2m_e \frac{hc}{\lambda}} \).

Step 3: Setup for New Momentum Condition

The problem asks for a new wavelength \( \lambda' \) such that the momentum is \( 1.5p \). Therefore, the new momentum equation is \( 1.5p = \sqrt{2m_e \frac{hc}{\lambda'}} \).

Step 4: Solving for New Wavelength

Set up the equation from the new momentum condition: \[ \left(1.5 \sqrt{2m_e \frac{hc}{\lambda}}\right)^2 = 2m_e \frac{hc}{\lambda'} \]. Simplifying gives: \[ 2.25 \cdot 2m_e \frac{hc}{\lambda} = 2m_e \frac{hc}{\lambda'} \].

Step 5: Simplification and Solution

Cancel out the common factors and solve for \( \lambda' \): \[ \frac{\lambda'}{\lambda} = \frac{1}{2.25} = \frac{4}{9} \]. Hence, \( \lambda' = \frac{4}{9} \lambda \).

Key Concepts

Photon Energy

The concept of photon energy is central to understanding how light interacts with matter. A single photon carries a specific amount of energy, which can be calculated using the equation:

• \( E = \frac{hc}{\lambda} \), where \( E \) is the energy, \( h \) is Planck's constant (approximately \( 6.626 \times 10^{-34} \) J·s), \( c \) is the speed of light (about \( 3 \times 10^8 \) m/s), and \( \lambda \) is the wavelength of the light.

This formula shows that photon energy is inversely related to its wavelength—the shorter the wavelength, the higher the energy of the photon. This relationship plays a crucial role in the photoelectric effect, where photons hit a metal surface and can eject electrons. Only photons with sufficient energy, greater than the work function of the metal, can dislodge electrons. The photon energy thus determines whether the ejection of electrons is possible and their kinetic energy upon ejection.

Electron Momentum

Momentum is a fundamental concept in physics, representing the product of an object's mass and velocity. For the electrons freed during the photoelectric effect, their momentum can be expressed with the formula:

• \( p = \sqrt{2m_e KE} \), where \( p \) is the momentum, \( m_e \) is the mass of the electron (about \( 9.109 \times 10^{-31} \) kg), and \( KE \) is the kinetic energy of the electron.

In the context of the photoelectric effect, the momentum of ejected electrons is directly linked to their kinetic energy, which itself depends on the energy of the incident photons. If we assume that the kinetic energy is much higher than the work function, this relation simplifies calculations by focusing mainly on the kinetic component. Increasing the photon energy increases the kinetic energy, thus increasing the electron's momentum. Therefore, changes in light's wavelength can lead to changes in the electron's momentum, affecting calculations like those shown in the exercise problem.

Kinetic Energy

Kinetic energy is the energy that an electron has due to its motion after being ejected from a metal surface in the photoelectric effect. The kinetic energy of an ejected electron can often be approximated by:

• \( KE \approx \frac{hc}{\lambda} \), assuming that the kinetic energy is much greater than the metal's work function.

This simplification allows us to focus on the relationship between the light's wavelength and the ejected electron's kinetic energy. Essentially, as the energy of incident photons increases (by decreasing wavelength), more energy is available to transfer as kinetic energy to the ejected electron. Solving problems like the one in the exercise requires understanding how the shifting balance of energy between incoming photons and outgoing electrons affects the outcome.

Work Function

The work function is a key concept when discussing the photoelectric effect. It is the minimum energy required to remove an electron from the surface of a metal. Each metal has a specific work function, which depends on the nature of the material and its surface properties. In photoelectric problems:

• If the photon's energy exceeds this work function, electrons are ejected.
• Any excess energy goes into the kinetic energy of the ejected electrons.

However, in problems such as the one given, where the kinetic energy is assumed to be much greater than the work function, the focus shifts almost entirely to the kinetic component. This allows us to ignore the role of the work function in calculations, simplifying the problem-solving process. Nonetheless, understanding the work function's importance helps in comprehending the threshold behavior observed in the photoelectric effect where below a certain frequency, no electrons are emitted, no matter the intensity of the light.