(b) A Free Electron Has A Wavefunction (x)=Acos[(10^10)xt], Where X Is In Meters. Find (i) The Electron's

(b) A Free Electron Has A Wavefunction (x)=Acos[(10^10)xt], Where x Is In Meters. Find (i) The Electron's

Understanding the wavefunction of a free electron is fundamental to quantum mechanics, as it encapsulates the probabilistic nature of electrons and their wave-like behavior. In this article, we will analyze the given wavefunction, determine key properties such as momentum, energy, and wavelength, and explore the significance of these quantities in the context of quantum physics. The wavefunction provided, \( \psi(x,t) = A \cos[(10^{10}) x t] \), offers a fascinating glimpse into the wave behavior of electrons and serves as an excellent example to illustrate core concepts.

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Introduction to Wavefunctions in Quantum Mechanics

Before delving into the specifics of the problem, it's essential to understand what a wavefunction represents in quantum mechanics.

Definition of a Wavefunction

• The wavefunction, typically denoted as \( \psi(x, t) \), contains all the information about a quantum particle.
• The probability density of finding a particle at position \( x \) at time \( t \) is given by \( |\psi(x,t)|^2 \).
• For a free electron, the wavefunction propagates without external potential influences, exhibiting wave-like properties such as interference and diffraction.

Mathematical Form of the Wavefunction

• The form \( \psi(x,t) = A \cos(k x - \omega t) \) is common for plane waves, where \( k \) is the wave number and \( \omega \) is the angular frequency.
• The given wavefunction is \( \psi(x,t) = A \cos[(10^{10}) x t] \), which suggests a wave-like behavior with specific parameters to be determined.

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Analyzing the Given Wavefunction

Let's examine the wavefunction:

\[
\psi(x,t) = A \cos[(10^{10}) x t]
\]

This form indicates a spatial and temporal dependence intertwined through the argument of the cosine function. To analyze the physical properties, we need to interpret the parameters correctly.

Identifying the Wave Parameters

• The argument of the cosine, \( (10^{10}) x t \), suggests a combined wave number and angular frequency.
• Typically, a plane wave is expressed as \( \psi(x,t) = A \cos(k x - \omega t) \), but here, the dependence is \( \cos(k x t) \).

This form is unusual because it involves a product \( x t \), which is non-standard. However, for the purpose of analysis, we can interpret this as indicating a phase that depends on \( x t \), implying a wave with properties that vary in a specific manner.

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Deriving Physical Quantities

To find key properties such as the momentum and energy of the electron, we need to relate the wavefunction to standard quantum mechanical expressions.

Momentum of the Electron

• In quantum mechanics, the momentum operator in the position representation is:

\[ \hat{p} = -i \hbar \frac{\partial}{\partial x} \]

• To find the expectation value of momentum, \( \langle p \rangle \), we typically calculate:

\[ \langle p \rangle = \int \psi^(x,t) \hat{p} \psi(x,t) dx \]

Given the form of \( \psi \), it's more straightforward to analyze the wavevector \( k \).

Note: The given wavefunction's argument involves \( x t \). To interpret this in standard form, consider the phase:

\[
\phi(x,t) = (10^{10}) x t
\]

which suggests a phase velocity or a wave vector that depends on \( t \). Alternatively, we can attempt to rewrite or approximate the wavefunction.

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Approximate or Reinterpret the Wavefunction

• If we treat the wavefunction as a standing wave or a wave with a time-dependent wave number, we might need to consider a different approach.
• However, for simplicity, assuming the wavefunction represents a traveling wave in some form, the wave number \( k \) and angular frequency \( \omega \) can be identified from the coefficients.

Given the standard form of a plane wave:

\[
\psi(x,t) = A \cos(k x - \omega t)
\]

and comparing it to:

\[
\psi(x,t) = A \cos[(10^{10}) x t]
\]

we see the phase depends on \( x t \), which is non-standard. But if we consider the phase as:

\[
\phi = (10^{10}) x t
\]

then the phase velocity \( v_p \) is:

\[
v_p = \frac{\omega}{k}
\]

where, assuming the phase velocity relates to the coefficients, we can identify:

\[
k = (10^{10}) t
\]
\[
\omega = (10^{10}) x
\]

which implies both \( k \) and \( \omega \) depend on \( t \) and \( x \), respectively. This suggests a more complex behavior, possibly a wave packet or a non-traveling wave.

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Standard Approach: Assuming a Plane Wave Form

Given the standard approach to such problems, we assume the wavefunction can be approximated as:

\[
\psi(x,t) = A \cos(k x - \omega t)
\]

with

\[
k = 10^{10} \text{ m}^{-1}
\]
\[
\omega = 10^{10} \text{ s}^{-1}
\]

This is a common assumption for plane waves with well-defined wave number and frequency.

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Calculating the Electron's Momentum

• The momentum \( p \) relates to the wave number \( k \):

\[ p = \hbar k \]

where \( \hbar \) is the reduced Planck's constant:

\[
\hbar \approx 1.055 \times 10^{-34} \text{ Js}
\]

Therefore,

\[
p = 1.055 \times 10^{-34} \times 10^{10} = 1.055 \times 10^{-24} \text{ kg·m/s}
\]

This is the magnitude of the electron's momentum associated with the wavefunction.

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Calculating the Electron's Energy

• The total energy \( E \) of a free electron with momentum \( p \) is given by the relativistic or non-relativistic relation.

Non-Relativistic Approximation:

\[
E = \frac{p^2}{2m}
\]

where \( m \) is the electron mass:

\[
m \approx 9.11 \times 10^{-31} \text{ kg}
\]

Plugging in the values:

\[
E = \frac{(1.055 \times 10^{-24})^2}{2 \times 9.11 \times 10^{-31}} \approx \frac{1.113 \times 10^{-48}}{1.822 \times 10^{-30}} \approx 6.11 \times 10^{-19} \text{ J}
\]

Expressed in electronvolts (eV), using \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \):

\[
E \approx \frac{6.11 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 3.81 \text{ eV}
\]

Thus, the electron's kinetic energy is approximately 3.81 eV.

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Calculating the Wavelength

• The de Broglie wavelength \( \lambda \) is related to the wave number \( k \):

\[ \lambda = \frac{2 \pi}{k} \]

Given \( k = 10^{10} \text{ m}^{-1} \):

\[
\lambda = \frac{2 \pi}{10^{10}} \approx \frac{6.2832}{10^{10}} \approx 6.2832 \times 10^{-10} \text{ meters}
\]

This wavelength (~0.63 nanometers) is typical of high-energy electrons.

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Summary of Results

| Quantity | Value | Explanation |
|--------------------------|-----------------------------------------------------|-----------------------------------------------------|
| Electron's momentum (\( p \)) | \( 1.055 \times 10^{-24} \text{ kg·m/s} \) | Derived from \( p = \hbar k \) |
| Electron's energy (\( E \)) | approximately \( 3.81 \text{ eV} \) | Non-relativistic kinetic energy |
| Wavelength (\( \lambda \)) | approximately \( 0.63 \text{ nm} \) | de Broglie wavelength associated with \( p \) |

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Significance of These Quantities in Quantum Mechanics

Understanding the momentum, energy, and wavelength of an electron described by a wavefunction like \(