The electrons in a metal that produce electric currents behave approximately as molecules of an ideal gas. The mass of an electron is \(m_{\mathrm{e}} \doteq 9.109 \cdot 10^{-31} \mathrm{~kg} .\) If the temperature of the metal is \(300.0 \mathrm{~K},\) what is the root-mean-square speed of the electrons?

Short Answer

Expert verified
Answer: The root-mean-square speed of the electrons in the metal is approximately \(1.079 \times 10^4~m/s\).

Step by step solution

01

Write down the given values

We have been given the mass of the electron (m) and the temperature (T) of the metal. Let's write them down: \(m_e = 9.109 \times 10^{-31}~kg\) \(T= 300.0~K\)
02

Find the Boltzmann constant k

The Boltzmann constant (k) is a physical constant that relates the average kinetic energy of particles in a gas with the temperature of the gas. Its value is: \(k = 1.381 \times 10^{-23} \frac{J}{K}\)
03

Use the formula of root-mean-square speed

We can now use the formula for the root-mean-square speed: \(v_{rms} = \sqrt{\frac{3kT}{m}}\) Plug in the given values: \(v_{rms} = \sqrt{\frac{3(1.381 \times 10^{-23} \frac{J}{K})(300.0 K)}{(9.109 \times 10^{-31} kg)}}\)
04

Calculate v_rms

Perform the calculations to find the root-mean-square speed of the electrons: \(v_{rms} = \sqrt{\frac{3(1.381 \times 10^{-23})(300.0)}{(9.109 \times 10^{-31})}}\) \(v_{rms} = \sqrt{1.164 \times 10^{8}}\) \(v_{rms} = 1.079 \times 10^4 m/s\) Therefore, the root-mean-square speed of the electrons in the metal is approximately \(1.079 \times 10^4~m/s\).

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Key Concepts

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

Boltzmann Constant
The Boltzmann constant (symbol: k or k_B) is a fundamental constant in physics that plays a key role in the kinetic theory of gases. It essentially bridges the gap between the macroscopic and microscopic worlds by connecting temperature, a macroscopic property, with the average kinetic energy of particles, a microscopic property.

Expressed in joules per kelvin (J/K), its value is approximately \(1.381 \times 10^{-23} J/K\). When studying the behavior of gases, this constant is used to calculate the average kinetic energy per particle with the formula \(\langle E_{\text{kin}} \rangle = \frac{3}{2} k_B T\), where \(T\) is the temperature in kelvins. In the context of root-mean-square speed for electrons, as in our exercise, the Boltzmann constant directly informs us of the energy associated with their thermal motion at a given temperature.
Electron Mass
Electrons are subatomic particles with a specific mass denoted by \(m_e\), which is essential for calculations in both atomic physics and solid state physics. The mass of an electron, \(9.109 \times 10^{-31} kg\), is a constant that allows us to compute various properties of electrons when coupled with other constants and equations.

Understanding electron mass is crucial when it comes to discussing the motion of electrons, such as their speed in a metal conductor. Here, it is important to note that the mass of an electron remains constant, regardless of its speed. This trait greatly simplifies many physics calculations, including our example where we determine the root-mean-square speed of electrons within the metal.
Kinetic Theory of Gases
The kinetic theory of gases is a theoretical framework that describes a gas as a collection of particles in random motion, colliding elastically with each other and with the walls of any container. One of the principal outcomes of this theory is that the temperature of a gas is directly proportional to the average kinetic energy of its molecules.

One key aspect of this theory is the concept of root-mean-square speed (\(v_{rms}\)). It is the measure of the speed of particles in a gas, which when squared and averaged, gives an indication of the kinetic energy in the system. It is calculated using the formula \(v_{rms} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of a single particle. In essence, the kinetic theory provides a molecular-level insight into the macroscopic properties of gases, which in turn can be applied to electrons in metals as in our exercise, considering them as particles in a 'gas' of electrons.
Thermal Physics
Thermal physics is the branch of physics that deals with heat, temperature, and their relation to energy, work, radiation, and the properties of matter. It encompasses the study of thermodynamics, kinetic theory, and statistical mechanics. A fundamental aspect of thermal physics is understanding how energy is transferred within a system and how it affects the system's properties.

If we consider the electrons in a metal as per our example, thermal physics would examine how the electrons' thermal energy (dictated by temperature) translates into their kinetic energy and subsequently their motion. By understanding these relationships, thermal physics allows us to predict how the electrons will behave under different thermal conditions, such as how fast they will move at a certain temperature, which is fundamentally related to the concept of root-mean-square speed we encountered in the exercise.

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Most popular questions from this chapter

The compression and rarefaction associated with a sound wave propogating in a gas are so much faster than the flow of heat in the gas that they can be treated as adiabatic processes. a) Find the speed of sound, \(v_{s}\), in an ideal gas of molar mass \(M\). b) In accord with Einstein's refinement of Newtonian mechanics, \(v_{\mathrm{s}}\) cannot exceed the speed of light in vacuum, \(c\). This fact implies a maximum temperature for an ideal gas. Find this temperature. c) Evaluate the maximum temperature of part (b) for monatomic hydrogen gas (H). d) What happens to the hydrogen at this maximum temperature?

Compare the average kinetic energy at room temperature of a nitrogen molecule to that of a nitrogen atom. Which has the larger kinetic energy? a) nitrogen atom b) nitrogen molecule c) They have the same energy. d) It depends upon the pressure.

A glass of water at room temperature is left on the kitchen counter overnight. In the morning, the amount of water in the glass is smaller due to evaporation. The water in the glass is below the boiling point, so how is it possible for some of the liquid water to have turned into a gas?

At a temperature of \(295 . \mathrm{K},\) the vapor pressure of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) is \(60.7 \mathrm{kPa}\). Suppose \(1.000 \mathrm{~g}\) of gaseous pentane is contained in a cylinder with diathermal (thermally conducting) walls and a piston to vary the volume. The initial volume is \(1.000 \mathrm{~L},\) and the piston is moved in slowly, keeping the temperature at \(295 \mathrm{~K}\). At what volume will the first drop of liquid pentane appear?

Calculate the change in internal energy of 1.00 mole of a diatomic ideal gas that starts at room temperature \((293 \mathrm{~K})\) when its temperature is increased by \(2.00 \mathrm{~K}\).

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