Problem 17 - Master Resource Book in JEE Main Physics

Question

Ultraviolet radiation of \(6.2 \mathrm{eV}\) falls on an aluminium surface (work function \(4.2 \mathrm{eV}\) ). The kinetic energy in joule of the fastest electron emitted is approximately. (a) \(3 \times 10^{-21}\) (b) \(3.2 \times 10^{-19}\) (c) \(3 \times 10^{-17}\) (d) \(3 \times 10^{-15}\)

Step-by-Step Solution

Answer

The kinetic energy of the fastest electron emitted is approximately \( 3.2 \times 10^{-19} \mathrm{J} \).

Step 1: Understand the Photoelectric Effect Equation

The photoelectric effect is described by the equation \( E_k = hu - \phi \), where \( E_k \) is the kinetic energy of the emitted electron, \( hu \) is the energy of the incoming photon, and \( \phi \) is the work function of the metal. In this problem, \( hu = 6.2 \mathrm{eV} \) and \( \phi = 4.2 \mathrm{eV} \).

Step 2: Calculate the Kinetic Energy in Electronvolts

Use the equation from Step 1 to find the kinetic energy: \( E_k = 6.2 \mathrm{eV} - 4.2 \mathrm{eV} = 2.0 \mathrm{eV} \). So, the kinetic energy of the emitted electron in electronvolts is \( 2.0 \mathrm{eV} \).

Step 3: Convert Electronvolts to Joules

The conversion factor from electronvolts to joules is \( 1 \mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J} \). Therefore, the kinetic energy of the electron is \( 2.0 \mathrm{eV} \times 1.6 \times 10^{-19} \mathrm{J/eV} = 3.2 \times 10^{-19} \mathrm{J} \).

Step 4: Identify the Correct Answer

Match the calculated kinetic energy \( 3.2 \times 10^{-19} \mathrm{J} \) with the given options. Option (b) \( 3.2 \times 10^{-19} \mathrm{J} \) is the correct answer.

Key Concepts

Ultraviolet Radiation

Ultraviolet (UV) radiation is a type of electromagnetic radiation that is not visible to the naked eye. It falls in the spectrum between visible light and X-rays, and has shorter wavelengths than visible light. This radiation is commonly known for its energetic photons, which makes it well-suited to study the photoelectric effect.

In physics, UV radiation is crucial because it can provide enough energy to emit electrons from a metal surface. The emission of electrons when UV light shines on a material is a fundamental property of the photoelectric effect. Being aware of the properties and effects of different frequencies of UV light helps understand how and why certain materials emit electrons when exposed to this radiation.

• Ultraviolet light has energy measured in electronvolts (eV).
• Understanding the interaction of UV light with materials is essential for applications such as solar energy and detecting UV-based phenomena.

Work Function

The work function, often symbolized as \( \phi \), is the minimum amount of energy needed to remove an electron from a solid to a point outside the solid itself. This is a crucial concept when you consider how materials respond to external radiant energy, like ultraviolet radiation.

The value of the work function depends on the material. For example, in the given exercise, aluminum has a work function of \( 4.2 \mathrm{eV} \). This means that any photon with energy greater than \( 4.2 \mathrm{eV} \) can cause the emission of electrons from the aluminum surface.

• The work function is typically measured in electronvolts (eV).
• The threshold level of photon energy needed to trigger the photoelectric effect varies between materials.

Kinetic Energy

Kinetic energy in the context of the photoelectric effect refers to the energy that the emitted electron gains after absorbing a photon. For an electron to be emitted, the incoming photon must have energy greater than the work function of the material.

Using the photoelectric equation \( E_k = hu - \phi \), we calculate \( E_k \), the kinetic energy of the emitted electron. In our exercise, where the energy of the incoming photon is \( 6.2 \mathrm{eV} \) and the work function is \( 4.2 \mathrm{eV} \), we derived a kinetic energy of \( 2.0 \mathrm{eV} \). This kinetic energy represents the energy available to the electron after overcoming the work function barrier.

• Kinetic energy of emitted electrons is calculated in electronvolts (eV).
• The calculation helps understand energy dynamics involved in the photoelectric effect.

Electronvolt to Joule Conversion

An electronvolt (eV) is a unit of energy that is widely used in atomic and particle physics, while a joule (J) is the SI unit of energy. To effectively understand and solve problems involving these concepts, conversion between these units is essential.

In the step-by-step solution, we converted the kinetic energy of an electron from electronvolts to joules. The conversion factor is \( 1 \mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J} \). Thus, \( 2.0 \mathrm{eV} \) of kinetic energy is converted to \( 3.2 \times 10^{-19} \mathrm{J} \). This enables comparison with scientific datasets that use different energy units.

• Always remember the conversion factor: \( 1 \mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J} \).
• Converting units helps align energy calculations with existing scientific standards.