Effect of Quadrupling the Accelerating Potential on the de Broglie Wavelength of an Electron

Correct Answer: The de Broglie wavelength becomes half of its initial value, or changes by a factor of 1/2.

Understanding the de Broglie Wavelength of an Accelerated Electron

According to the de Broglie hypothesis, every moving material particle has an associated wavelength known as the de Broglie wavelength. For an electron, this wavelength depends on its momentum. When an electron is accelerated through a potential difference, it gains kinetic energy, which increases its momentum and consequently decreases its de Broglie wavelength.

The de Broglie wavelength is given by:

λ = h/p

Here, λ is the de Broglie wavelength, h is Planck’s constant, and p is the momentum of the electron. To determine how the wavelength changes when the accelerating potential is quadrupled, we need to establish the relationship between electron momentum and accelerating potential.

Relationship Between Accelerating Potential and Electron Kinetic Energy

When an electron of charge e is accelerated from rest through a potential difference V, the electrical energy gained by the electron is converted into kinetic energy. Therefore:

eV = (1/2)mv²

Here, e is the magnitude of the charge of the electron, V is the accelerating potential difference, m is the mass of the electron, and v is its speed after acceleration.

Since the momentum of the electron is:

p = mv

The kinetic energy can also be expressed in terms of momentum as:

K = p²/(2m)

Therefore:

eV = p²/(2m)

Rearranging the equation gives:

p = √(2meV)

Deriving the Relationship Between de Broglie Wavelength and Potential Difference

Substituting the momentum of the accelerated electron into the de Broglie relation gives:

λ = h/√(2meV)

For an electron, h, m, and e are constants. Therefore, the de Broglie wavelength depends only on the accelerating potential V according to the following proportionality:

λ ∝ 1/√V

This inverse square-root relationship is the key concept required to solve the question. It shows that increasing the accelerating potential decreases the de Broglie wavelength, but the wavelength does not decrease in direct proportion to the potential.

Calculating the Change in de Broglie Wavelength

Initial and Final Accelerating Potentials

Let the initial accelerating potential difference be V₁. The potential difference is then quadrupled, so the final potential difference becomes:

V₂ = 4V₁

Let the corresponding initial and final de Broglie wavelengths be λ₁ and λ₂, respectively.

Applying the Inverse Square-Root Relationship

Since the de Broglie wavelength is inversely proportional to the square root of the accelerating potential:

λ₂/λ₁ = √(V₁/V₂)

Substituting V₂ = 4V₁:

λ₂/λ₁ = √[V₁/(4V₁)]

Cancelling V₁:

λ₂/λ₁ = √(1/4)

Therefore:

λ₂/λ₁ = 1/2

Hence:

λ₂ = λ₁/2

Why Does the Wavelength Become Half?

Quadrupling the accelerating potential makes the kinetic energy of the electron four times greater because the kinetic energy gained by the electron is directly proportional to the potential difference. Since momentum is proportional to the square root of kinetic energy, the momentum of the electron becomes twice its original value.

The de Broglie wavelength is inversely proportional to momentum. Therefore, when the electron momentum doubles, its wavelength becomes half of the original wavelength. This explains physically why a fourfold increase in accelerating potential results in only a twofold decrease in the de Broglie wavelength.

General Relationship Between Potential and Electron Wavelength

For an electron accelerated through a potential difference V, the de Broglie wavelength follows the relationship λ ∝ 1/√V. Therefore, if the potential is multiplied by any factor n, the wavelength is divided by √n.

In this question, n = 4. Therefore, the wavelength is divided by √4 = 2. This gives the final wavelength as one-half of the initial wavelength.

Final Answer

When the accelerating potential difference is quadrupled, the de Broglie wavelength of the electron beam becomes half of its initial value.

Therefore, the de Broglie wavelength changes by a factor of 1/2.