Question

A proton accelerated through a potential difference V has a de Broglie wavelength λ. On doubling the accelerating potential, the de Broglie wavelength of the proton_____.
Fill in the blank with the correct answer from the options given below

• A. Remains unchanged
• B. Becomes double
• C. Becomes four times
• D. Decreases

Answer 1

The Correct Option is D

Let's analyze the relationship between the de Broglie wavelength and the accelerating potential.

1. De Broglie Wavelength:

The de Broglie wavelength (λ) is given by:

λ = h / p

Where:

• h is Planck's constant
• p is the momentum of the particle

2. Kinetic Energy and Momentum:

When a proton is accelerated through a potential difference V, its kinetic energy (KE) is given by:

KE = qV

Where:

• q is the charge of the proton

The kinetic energy is also related to momentum (p) by:

KE = p² / (2m)

Where:

• m is the mass of the proton

Therefore:

qV = p² / (2m)

p² = 2mqV

p = √(2mqV)

3. De Broglie Wavelength in terms of Potential:

Substituting p into the de Broglie wavelength equation:

λ = h / √(2mqV)

4. Effect of Doubling the Potential:

Let's consider the initial wavelength (λ₁) when the potential is V:

λ₁ = h / √(2mqV)

Now, let's consider the wavelength (λ₂) when the potential is doubled (2V):

λ₂ = h / √(2mq(2V))

λ₂ = h / √(4mqV)

λ₂ = h / (√(2)√(2mqV))

λ₂ = λ₁ / √2

Therefore, when the accelerating potential is doubled, the de Broglie wavelength decreases by a factor of √2.

The correct answer is:

Option 4: Decreases