Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

What is the work function of the metal if the light of wavelength \(4000 \AA\) generates photoelectrons of velocity \(6 \times 10^{5} \mathrm{~ms}^{-1}\) from it ? (Mass of electron= \(9 \times 10^{-31} \mathrm{~kg}\) Velocity of light \(=3 \times 10^{\circ} \mathrm{ms}^{-1}\) Planck's constant \(=6.626 \times 10^{-34} \mathrm{~J}_{\mathrm{S}}\) Charge of electron \(=1.6 \times 10^{-19} \mathrm{JeV}^{-1}\) ) (a) \(0.9 \mathrm{eV}\) (b) \(3.1 \mathrm{eV}\) (c) \(2.1 \mathrm{eV}\) (d) \(4.0 \mathrm{eV}\)

Step-by-Step Solution

Answer

The work function is 2.1 eV, which matches option (c).

Step 1: Convert Wavelength to Meters

The given wavelength of light is \(4000 \AA\). To convert this into meters, use the conversion \(1 \AA = 10^{-10} \text{ meters}\). Thus, the wavelength \(\lambda = 4000 \times 10^{-10} \text{ meters} = 4 \times 10^{-7} \text{ meters}\).

Step 2: Calculate the Energy of Incident Light

Use the formula for energy of a photon: \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34} \text{ Js}\), \(c = 3 \times 10^{8} \text{ m/s}\), and \(\lambda = 4 \times 10^{-7} \text{ meters}\). Calculate as follows:\[E = \frac{(6.626 \times 10^{-34}) (3 \times 10^{8})}{4 \times 10^{-7}} = 4.9695 \times 10^{-19} \text{ Joules}\].

Step 3: Calculate Kinetic Energy of Emitted Photoelectrons

The kinetic energy \(KE\) of the photoelectrons can be calculated using \(KE = \frac{1}{2}mv^2\) where \(m = 9 \times 10^{-31} \text{ kg}\) and \(v = 6 \times 10^{5} \text{ m/s}\). Thus:\[KE = \frac{1}{2} (9 \times 10^{-31}) (6 \times 10^{5})^2 = 1.62 \times 10^{-19} \text{ Joules}\].

Step 4: Apply Photoelectric Equation

Use the photoelectric equation: \(E = \Phi + KE_{max}\) where \(\Phi\) is the work function. Rearrange for \(\Phi\):\[\Phi = E - KE = 4.9695 \times 10^{-19} - 1.62 \times 10^{-19} = 3.3495 \times 10^{-19} \text{ Joules}\].

Step 5: Convert Work Function to Electron Volts

To convert the work function into electron volts, use the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ Joules}\). Thus:\[\Phi (\text{in eV}) = \frac{3.3495 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0934375 \approx 2.1 \text{ eV}\].

Step 6: Identify the Correct Option

Compare \(\Phi \approx 2.1 \text{ eV}\) to the options provided. The closest matching option is \(\text{(c) } 2.1 \text{ eV}\).

Key Concepts

Work Function

The work function, represented by the Greek letter \( \Phi \), is an essential concept in the photoelectric effect. It is defined as the minimum energy required to remove an electron from the surface of a material, usually a metal. In the context of the photoelectric effect, this energy is supplied by a photon. The work function is a characteristic of the material and can vary greatly depending on the type of metal.
The work function can be determined by the equation:

• \( E = \Phi + KE_{max} \)

Here, \( E \) is the energy of the incident photon and \( KE_{max} \) stands for the maximum kinetic energy of the emitted photoelectrons. By rearranging the formula, we obtain:

• \( \Phi = E - KE_{max} \)

In solving problems involving the work function, converting the energy into electron volts (eV) is often necessary since energy values in this unit are more intuitive for working with electrons.
This concept is crucial in understanding how light can eject electrons from a material, ultimately shedding light on the wave-particle duality of light.

Kinetic Energy of Photoelectrons

The kinetic energy of photoelectrons, denoted as \( KE \), is the energy that the ejected electrons possess after absorbing energy from an incident photon. This energy is governed by the equation:

• \( KE = \frac{1}{2} mv^2 \)

In this formula, \( m \) represents the mass of the electron, and \( v \) is its velocity. It helps to quantify how fast and energetically the electrons are moving after they have been freed from a metal's surface by incoming photons.
To calculate this, you need to know the velocity of the electrons, which is often provided or can be inferred from experiments. The kinetic energy tells us how much leftover energy the electrons have after overcoming their binding energy (work function) from the nucleus. This energy is central to applications like photoelectric sensors and understanding fundamental quantum effects on electron behavior.
Through understanding \( KE \), we grasp how energy is transferred from photons to electrons, illustrating one of the fascinating aspects of particle physics.

Photon Energy Calculation

The energy of a photon is a fundamental tool in the study of light-matter interaction, often calculated using the formula:

• \( E = \frac{hc}{\lambda} \)

Here, \( h \) represents Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon. This equation illustrates how the energy of a photon is inversely proportional to its wavelength, meaning shorter wavelengths have higher energy.
This relation is integral to understanding how photons of different energies interact with materials. For example, in the photoelectric effect, only photons with energy greater than or equal to the material's work function can eject electrons. By calculating photon energy, we establish the minimum requirements needed to achieve specific interactions.
Understanding photon energy provides insight into various technologies, including solar cells and lasers, as well as foundational theories in quantum mechanics.