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The de Broglie wavelength of a gas molecule at a temperature \(T \mathrm{~K}\) is (a) \(\frac{h}{\sqrt{3 m K T}}\) (b) \(\frac{h}{3 m K T}\) (c) \(\frac{h}{\sqrt{2 m K T}}\) (d) \(\sqrt{2 m K T}\)

Short Answer

Expert verified
The correct option is (a): \( \frac{h}{\sqrt{3 m K T}} \).

Step by step solution

01

Identify Given Parameters

The de Broglie wavelength is related to the momentum of a particle. To find the de Broglie wavelength of a gas molecule at a certain temperature, we need to find the expression that relates kinetic energy to temperature. The kinetic energy of a gas molecule is usually given by \( \frac{3}{2} k T \), where \( k \) is the Boltzmann constant.
02

Relate Kinetic Energy to Velocity

The kinetic energy of a molecule is also given by \( \frac{1}{2} m v^2 \) where \( m \) is the mass of the gas molecule and \( v \) is the velocity. Set \( \frac{1}{2} mv^2 = \frac{3}{2} k T \) to relate velocity to temperature. Solving for velocity, \( v = \sqrt{\frac{3kT}{m}} \).
03

Calculate Momentum

Momentum \( p \) is given by the product of mass and velocity, so \( p = m v \). Substituting the expression for \( v \), we have \( p = m \sqrt{\frac{3kT}{m}} = \sqrt{3m k T} \).
04

Apply de Broglie Wavelength Formula

The de Broglie wavelength \( \lambda \) is given by \( \lambda = \frac{h}{p} \). Substitute \( p = \sqrt{3m k T} \) into this expression: \( \lambda = \frac{h}{\sqrt{3m k T}} \).
05

Select the Correct Option

Comparing the derived expression \( \frac{h}{\sqrt{3 m K T}} \) with the options provided, the correct answer is option (a): \( \frac{h}{\sqrt{3 m K T}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is one of the fundamental concepts in physics related to how objects move. It's the energy that a particle or an object has due to its motion. Quite simply, it's expressed as
  • the amount of work required to accelerate an object from rest to its current velocity.
  • quantified by the formula: \( KE = \frac{1}{2} mv^2 \) where \( m \) is the mass and \( v \) is the velocity of the object.
When it comes to gases and their molecules, things get even more interesting. Here, kinetic energy is often tied to temperature. In the realm of gases, we use the concept of temperature to talk about the average kinetic energy of molecules. That's where the equation \( KE = \frac{3}{2} k T \) comes in, linking kinetic energy with temperature through the Boltzmann constant \( k \). This looks at energy per molecule rather than for large objects.

Understanding kinetic energy in terms of both formulae is crucial when determining the behavior of particles, especially when looking into concepts like the de Broglie wavelength. In doing so, finding velocity becomes an important bridge between kinetic energy and momentum.
Momentum
Momentum is all about the quantity of motion an object possesses. It's defined as the product of mass and velocity: \( p = mv \). Essentially, it tells us how much motion an object is carrying and how hard it would be to stop it. For a gas molecule, we aren't typically concerned with stopping it, but rather how its momentum interacts with other particles or boundaries.

This concept of momentum is crucial when looking at de Broglie's hypothesis, which combines aspects of wave and particle theories. To find the de Broglie wavelength, we use the formula \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is momentum.
  • To apply this, we first relate kinetic energy to velocity, leading to \( v = \sqrt{\frac{3kT}{m}} \).
  • Thus, momentum \( p \) becomes \( m \sqrt{\frac{3kT}{m}} = \sqrt{3mkT} \).
So, understanding momentum not only helps in solving physics problems, but in seeing how particle's environments, like temperature, influence their behavior.
Boltzmann Constant
The Boltzmann constant, symbolized by \( k \), acts as a bridge between macroscopic and microscopic physical quantities. It's a fundamental physical constant that links temperature with energy at the individual particle level.

The Boltzmann constant appears in several pivotal physics equations:
  • In the ideal gas law as \( PV = NkT \), it relates volume, pressure, and temperature for an ideal gas.
  • In thermal energy calculations, such as \( KE = \frac{3}{2}kT \), revealing how temperature correlates with the kinetic energy of particles.
When considering phenomena like the de Broglie wavelength, the Boltzmann constant helps relate the thermal motion and kinetic properties of particles. By characterizing energy per particle, we use \( k \) to explain processes rooted deep in thermodynamics and statistical mechanics.

Its value, \( 1.38 \times 10^{-23} J/K \), allows us to unify these spaces and gain deeper insight into the workings of molecules, especially in varied temperatures.
Temperature
Temperature is a familiar concept, often linked with how hot or cold something feels. In the scientific realm, temperature has a precise significance. It's a measure of the average kinetic energy of the particles in a substance. Instead of being just about warmth, it's fundamentally about energy motion.
  • An increase in temperature implies more energetic particles, moving faster and colliding with greater force.
  • Scientifically, temperature affects how molecules in gases behave, influencing both kinetic energy and momentum.
Temperature itself is measured in Kelvin (K) within scientific studies, providing a clear and consistent method to evaluate thermal energy.

Through concepts like the de Broglie wavelength for gases, temperature becomes essential in determining the wavelength of these molecules, reflecting how quickly they move at certain thermal conditions. With rising temperatures, not only does the energy increase, but calculations based on temperature also illustrate changes in molecular behaviors, seen through statistical mechanics and gas laws.

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Most popular questions from this chapter

Photoelectric effect is the phenomenon in which (a) photons come out of a metal when it is hit by a beam of electrons (b) photons come out of the nucleus of an atom under the action of an electric field (c) electrons come out of a metal with a constant velocity which depends upon the frequency and intensity of incident light wave (d) electrons come out of a metal with different velocities not greater than a certain value which depends only on the frequency of the incident light wave and not on its intensity

A light source is at a distance \(d\) from a photoelectric cell, then the number of photoelectrons emitted from the cell is \(n .\) If the distance of light source and cell is reduced to half, then the number of photoelectrons emitted will become (a) \(\frac{n}{2}\) (b) \(2 n\) (c) \(4 n\) (d) \(n\)

Light of wavelength \(0.6 \mu \mathrm{m}\) from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is \(0.5\) volt. With light of wavelength \(0.40 \mu \mathrm{m}\) from a mercury vapour lamp the stopping potential is \(1.5\) volt; then the work function in electron volts of the photocell surface is (a) \(0.75 \mathrm{eV}\) (b) \(1.5 \mathrm{eV}\) (c) \(3 \mathrm{eV}\) (d) \(2.5 \mathrm{eV}\)

\(X\) -rays are diffracted from a crystal of lattice plane spacing \(2 \AA\). The maximum wavelength that can be diffracted is (a) \(1 \AA\) (b) \(2 \AA\) (c) \(2.5 \AA\) (d) \(4 \AA\)

According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal versus frequency of the incident radiation gives a straight line whose slope (a) depends on the nature of metal used (b) depends on the intensity of radiation (c) depends on both the intensity of radiation and the nature of metal used (d) is the same for all metals and independent of the intensity of radiation

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