Problem 9 - Master Resource Book in JEE Main Physics

Question

Consider a metal exposed to light of wavelength \(600 \mathrm{~nm}\). The maximum energy of the electron doubles when light of wavelength \(400 \mathrm{~nm}\) is used. The work function in eV is \(\quad\) [NCERT Exemplar] (a) \(1.50 \mathrm{eV}\) (b) \(1.02 \mathrm{eV}\) (c) \(1.94 \mathrm{eV}\) (d) \(2.76 \mathrm{eV}\)

Step-by-Step Solution

Answer

The work function is approximately 1.02 eV, so the answer is (b).

Step 1: Understanding the Problem

In this problem, we're dealing with photoelectric effect. A metal emits electrons when exposed to light of certain wavelength, and we are asked to find the work function using two different wavelengths. The energy of the emitted electron is doubled when the wavelength changes from \(600 \text{ nm}\) to \(400 \text{ nm}\).

Step 2: Using Energy and Wavelength Formula

The energy of a photon is given by \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. We use \(E_1\) and \(E_2\) to denote energies for wavelengths \(600\,\text{nm}\) and \(400\,\text{nm}\), respectively: \[E_1 = \frac{hc}{600 \, \text{nm}}\] \[E_2 = 2 \cdot E_1 = \frac{hc}{400 \, \text{nm}}\]

Step 3: Setting Up the Photon Energy Equations

Express the relation between photon energy, maximum kinetic energy of electrons, and the work function (\(\phi\)) as \(E_1 - \phi = K_1\) and \(E_2 - \phi = K_2 = 2K_1\). Simplifying using the energy equations, we get:\[ \frac{hc}{600 \, \text{nm}} - \phi = K_1 \]\[ \frac{hc}{400 \, \text{nm}} - \phi = 2K_1 \]

Step 4: Eliminate Kinetic Energy

From the equation in Step 3, equate the kinetic energy expressions to solve for \(\phi\):\[ \frac{hc}{600 \, \text{nm}} - \phi = \frac{1}{2} \left( \frac{hc}{400 \, \text{nm}} - \phi \right) \]Rearrange and solve:\[ 2 \left( \frac{hc}{600 \, \text{nm}} \right) - 2\phi = \frac{hc}{400 \, \text{nm}} - \phi \]

Step 5: Solve for Work Function \(\phi\)

After rearranging, we have:\[ 2\phi = \frac{hc}{400 \, \text{nm}} - 2\frac{hc}{600 \, \text{nm}} \]\[ \phi = \frac{hc}{400 \, \text{nm}} - \frac{hc}{600 \, \text{nm}} \]Substitute \(h = 4.1357 \times 10^{-15} \text{ eV}\cdot \text{s}\) and \(c = 3 \times 10^8 \text{ m/s}\):Calculate each term to find \(\phi\):\(\frac{hc}{400 \text{ nm}} = \frac{4.1357 \times 10^{-15} \times 3 \times 10^8}{400} \text{ eV} \)Similarly:\(\frac{hc}{600 \text{ nm}} = \frac{4.1357 \times 10^{-15} \times 3 \times 10^8}{600} \text{ eV} \)Finally, subtract these:\[\phi = (3.10 - 2.07) \text{ eV} = 1.03 \text{ eV} \]

Step 6: Select Correct Answer

The calculated work function \(1.03 \text{ eV}\) is the closest to option (b) \(1.02 \text{ eV}\). Thus, the correct answer is (b).

Key Concepts

Work Function

In the context of the photoelectric effect, the work function, denoted by \(\phi\), is a critical concept. It represents the minimum energy required to dislodge an electron from the surface of a metal. Each metal has a unique work function depending on how strongly it holds onto its electrons.
For the electron emission to occur, the energy provided by the incident photon must at least equal this work function. If the photon's energy exceeds the work function, the excess energy transforms into the kinetic energy of the emitted electron.
In the given exercise, knowing the work function helps us understand why an electron can be emitted at different energy levels when the incident light's wavelength changes.

Planck's Constant

Planck's constant \(h\) is a fundamental constant in physics, pivotal in the study of quantum mechanics. It quantifies the action present in any quantum event.
Planck's constant links the energy of a photon to its frequency, expressed in the formula \(E = h \cdot f\), where \(E\) is energy and \(f\) is frequency. For visible light, it's more common to express energy in terms of wavelength using \(E = \frac{hc}{\lambda}\), where \(c\) is the speed of light.
In the exercise, Planck's constant allows us to compute the energy possessed by photons at given wavelengths, which is fundamental in determining the work function and the energy of emitted electrons from a metal surface.

Photon Energy

Photon energy refers to the energy carried by a single photon, closely linked with its wavelength or frequency. The energy can be computed with \(E = \frac{hc}{\lambda}\). This expression reveals that photon's energy is inversely proportional to the wavelength. A shorter wavelength indicates higher energy, as seen in the exercise when transitioning from 600 nm to 400 nm.
Understanding photon energy is crucial when discussing phenomena such as the photoelectric effect, where incoming light supplies the required energy to release electrons from a metal's surface.

• For 600 nm, the photon energy is lower compared to 400 nm.
• The change in energies illustrates why the maximum electron kinetic energy doubles with the shift in wavelength.

Electron Emission

Electron emission is the process where electrons are liberated from a solid animated by energy absorption, typically from photons in the photoelectric effect. When photons interact with a metal surface and transfer sufficient energy, electrons absorb this energy and overcome the work function barrier, leading to their emission.
The kinetic energy of these emitted electrons can be measured and is represented as the difference between the incident photon's energy and the metal's work function.
This concept is the driving force behind technologies like photoelectric cells, where light is converted into electrical energy by capturing the kinetic energy of emitted electrons. In the given problem, understanding changes in electron emission helps predict how alterations in light properties affect the emission efficiency and electron kinetic energy.