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

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

\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}} \]

where \( h \) is Planck's constant, \( p \) is the momentum, \( m \) is the mass of the particle, and \( E_k \) is the kinetic energy.

For a proton accelerated through a potential difference \( V \), the kinetic energy can be expressed as:

\[ E_k = eV \]

where \( e \) is the elementary charge.

Substituting \( E_k \) in the de Broglie equation:

\[ \lambda = \frac{h}{\sqrt{2meV}} \]

On doubling the accelerating potential to \( 2V \), the new wavelength \( \lambda' \) is given by:

\[ \lambda' = \frac{h}{\sqrt{2m \cdot 2eV}} = \frac{h}{\sqrt{4meV}} \]

This simplifies to:

\[ \lambda' = \frac{h}{2\sqrt{meV}} = \frac{\lambda}{\sqrt{2}} \]

Thus, the de Broglie wavelength decreases on doubling the accelerating potential. Therefore, the correct answer is "Decreases".