Question

The de - Broglie wavelength of a particle of kinetic energy 'K' is \(\lambda\), the wavelength of the particle if its kinetic energy is

• A. \(\frac{\lambda}{2}\)
• B. \(\lambda\)
• C. \(4\lambda\)
• D. \(2\lambda\)

Answer 1

The Correct Option is D

\( \lambda = \frac{h}{p} \)

Where:

• \(\lambda\) = de Broglie wavelength 
• h = Planck's constant
• p = momentum of the particle

The momentum of the particle can be related to its kinetic energy (\( K \)) as:

\( p = \sqrt{2mK} \)

Where:

• m = mass of the particle
• K = kinetic energy of the particle

Now, consider the situation where the kinetic energy of the particle is: \( K' = \frac{\lambda}{4} \)

We can calculate the new wavelength (\( \lambda' \)) using the same equation:

\( \lambda' = \frac{h}{p'} \)

Substitute the expression for momentum (\( p' \)) in terms of the new kinetic energy (\( K' \)):

\( \lambda' = \frac{h}{\sqrt{2mK'}} \)

Substitute \( K' = \frac{\lambda}{4} \) into the above equation:

\( \lambda' = \frac{h}{\sqrt{2m \left( \frac{\lambda}{4} \right)}} \)

Simplifying the expression:

\( \lambda' = \frac{h}{\sqrt{2m\frac{\lambda}{8}}} = \frac{h}{\left(\sqrt{\frac{\lambda}{4}} \times \sqrt{2m}\right)} = \frac{2h}{\sqrt{\lambda \times 2m}} \)

We can see that the new wavelength \( \lambda' \) is equal to:

\( \lambda' = 2\lambda \)

Therefore, the correct option is: (D) \( 2\lambda \)