Find the de broglie wavelength associated with an electron accelerated through a potential difference V.

To find the de Broglie wavelength associated with an electron that has been accelerated through a potential difference \( V \), we can use a combination of concepts from quantum mechanics and classical physics. The de Broglie wavelength is a fundamental concept that relates the wave-like properties of particles to their momentum. Let's break this down step by step.

Understanding the Basics

The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:

λ = h / p

where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)) and \( p \) is the momentum of the particle. For an electron, the momentum can be expressed in terms of its mass and velocity.

Calculating the Momentum

When an electron is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field. The kinetic energy (\( KE \)) gained by the electron can be expressed as:

KE = eV

where \( e \) is the charge of the electron (\( 1.602 \times 10^{-19} \, \text{C} \)). This kinetic energy can also be expressed in terms of momentum:

KE = p² / (2m)

Here, \( m \) is the mass of the electron (\( 9.109 \times 10^{-31} \, \text{kg} \)). Setting these two expressions for kinetic energy equal gives us:

eV = p² / (2m)

Solving for Momentum

Rearranging this equation to solve for momentum \( p \) yields:

p = √(2meV)

Finding the De Broglie Wavelength

Now that we have an expression for momentum, we can substitute it back into the de Broglie wavelength formula:

λ = h / p = h / √(2meV)

Substituting the known values:

• Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
• Mass of the electron \( m = 9.109 \times 10^{-31} \, \text{kg} \)
• Charge of the electron \( e = 1.602 \times 10^{-19} \, \text{C} \)

Final Expression

Thus, the de Broglie wavelength of the electron can be expressed as:

λ = h / √(2meV)

Example Calculation

Let’s say we want to find the de Broglie wavelength of an electron accelerated through a potential difference of \( 100 \, \text{V} \). Plugging in the values:

λ = 6.626 × 10⁻³⁴ / √(2 × 9.109 × 10⁻³¹ × 1.602 × 10⁻¹⁹ × 100)

Calculating the denominator:

√(2 × 9.109 × 10⁻³¹ × 1.602 × 10⁻¹⁹ × 100) ≈ 5.344 × 10⁻²²

Now, substituting this back into the wavelength equation gives:

λ ≈ 6.626 × 10⁻³⁴ / 5.344 × 10⁻²² ≈ 1.24 × 10⁻¹² \, \text{m}

Conclusion

So, the de Broglie wavelength of an electron accelerated through a potential difference of \( 100 \, \text{V} \) is approximately \( 1.24 \, \text{pm} \) (picometers). This illustrates how quantum mechanics allows us to understand the wave-like behavior of particles, even at the scale of electrons. If you have any further questions or need clarification on any part of this process, feel free to ask!