Problem 37 - Arihant AIEEE Physics

Question

The temperature at which average translational K.E. of molecule is equal to the K.E. of an electron accelerated from rest through a potential difference of \(1 \mathrm{~V}\), is : (a) \(T=7729 \mathrm{~K}\) (b) \(T=8879 \mathrm{~K}\) (c) \(T=7.72 \mathrm{~K}\) (d) \(T=772.9 \mathrm{~K}\)

Step-by-Step Solution

Answer

The correct answer is (a) \( T = 7729 \text{ K} \).

Step 1: Understanding the Electron Kinetic Energy

When an electron is accelerated from rest through a potential difference of 1 V, it gains kinetic energy. The kinetic energy (K.E.) gained by the electron is given by \( \text{K.E.} = eV \), where \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \text{ C} \)), and \( V \) is the potential difference (1 V). Hence, \( \text{K.E.} = 1.6 \times 10^{-19} \text{ J} \).

Step 2: Average Translational Kinetic Energy of a Molecule

According to kinetic theory, the average translational kinetic energy of a gas molecule is given by \( \frac{3}{2} k T \), where \( k \) is Boltzmann's constant \( (1.38 \times 10^{-23} \text{ J/K}) \) and \( T \) is the temperature in Kelvin.

Step 3: Equate the Kinetic Energies

We need to find the temperature \( T \) at which the average translational kinetic energy of a gas molecule equals the kinetic energy of the electron. So, equate \( \frac{3}{2} k T = 1.6 \times 10^{-19} \).

Step 4: Solving for Temperature \( T \)

Rearrange the equation from Step 3 to solve for \( T \): \[ T = \frac{2 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}} \] Calculate \( T \): \[ T = \frac{3.2 \times 10^{-19}}{4.14 \times 10^{-23}} \approx 7729 \text{ K} \]

Step 5: Conclusion and Answer Selection

Based on the calculations, the temperature at which the average translational kinetic energy of the molecule equals the kinetic energy of the electron is approximately 7729 K. Thus, the correct choice is (a) \( T = 7729 \text{ K} \).

Key Concepts

Average Kinetic Energy

In the world of gases, understanding the motion of molecules is key. Each molecule in a gas moves randomly and at different speeds. However, there's a neat way to summarize all this motion: average kinetic energy. It's sort of a way to think about the 'average speed' all those molecules are traveling.

Kinetic energy relates to an object's motion. For gas molecules, as they move faster due to an increase in temperature, their kinetic energy increases too. This is described by the formula \( \frac{3}{2} kT \), where:

• \(k\) is Boltzmann's constant.
• \(T\) is the temperature in Kelvin.

This average kinetic energy tells us about the temperature of the gas. Higher temperatures mean higher average kinetic energy as molecules move more vigorously.

So, when we solve for kinetic energy, we're looking for the temperature that gives the gas molecules an average speed or energy, much like when finding the speed limit for all cars on a freeway. It helps us to predict how gases will behave under different conditions.

Electron Kinetic Energy

Electrons are tiny particles within atoms that can gain energy when exposed to an electric field. When an electron is accelerated from rest, say through a potential difference of 1 volt, it gains kinetic energy. This gained energy can be thought of like the sprinting speed an athlete gets after a race start.

The kinetic energy acquired by an electron moving through a potential difference is given by \( eV \). Here:

• \(e\) is the electron's charge, \(1.6 \times 10^{-19} \text{ C}\).
• \(V\) is the potential difference, in this case, 1 volt.

Thus, the kinetic energy (K.E.) is calculated as simply \(1.6 \times 10^{-19} \text{ Joules}\). This amount is tiny compared to the everyday energies, but crucial at the atomic scale.

Understanding this concept helps us grasp how fundamental particles behave and interact with electric fields, altering their energy and motion.

Boltzmann's Constant

Boltzmann's constant, \( k \), is a fundamental constant that traces back to the microscopic nature of thermodynamics. It's like a bridge connecting the macroscopic world of temperature and the microscopic world of molecular motion.

With a value of \(1.38 \times 10^{-23} \text{ J/K} \), Boltzmann's constant is the key to unlocking the average kinetic energy of particles at a given temperature. This small number may seem negligible, but in calculations, it gives the required balance for the units of energy.

Boltzmann's constant appears in many fundamental formulas within physics, particularly in statistical mechanics, where it aids in expressing the energy distribution and entropy of a system. It's pivotal for translating the thermal behavior of molecules into understandable quantities.

Whether you're calculating the average kinetic energy of molecules or predicting the temperature scale in thermodynamics, Boltzmann's constant is always on the sidelines ensuring the measurements make sense.

Temperature in Kelvin

Temperature is a fundamental aspect of thermodynamics and kinetic theory. However, the Kelvin scale is the preferred choice for scientists studying such topics. Why? Because it starts at absolute zero.

Absolute zero is the point where molecular motion stops entirely, sitting at 0 Kelvin. It makes Kelvin a direct scale to measure thermal energy. Unlike Celsius or Fahrenheit, all temperature laws and equations in physics become simpler and more accurate using Kelvin.

When determining kinetic energies, scientists often refer to Kelvin. This scale helps directly relate temperature to energy through the formula for average kinetic energy \( \frac{3}{2} kT \).

Think of the Kelvin scale as a clear way to speak about energy - it's like using a universal language when discussing temperature and motion at the atomic level. It keeps the calculations consistent and allows scientists to predict molecule behavior reliably.