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

Answer: (D)

$2\lambda$

Answer 2

The de Broglie wavelength $\lambda$ of a particle is given by the equation:
$\lambda = \frac{h}{p}$
where λ is the wavelength, h is the Planck constant, and p is the momentum of the particle.

The momentum of a particle can be related to its kinetic energy (K) as:
$p = \sqrt{2mK}$ where m is the mass of the particle.
Now, let's consider the situation where the kinetic energy of the particle is $\frac{\lambda}{4}$
We can calculate the new wavelength (λ') using the same equation:
$\lambda' = \frac{h}{p'}$ where p' is the momentum of the particle with the new kinetic energy.

Substituting the expression for momentum p' in terms of K, we have:
$\lambda' = \frac{h}{\sqrt{2mK'}}$ where K' is the new kinetic energy, which is λ/4.
$\lambda' = \frac{h}{\sqrt{2m \left( \frac{\lambda}{4} \right)}}$
Simplifying the expression:
$\lambda' = \frac{h}{\sqrt{2m\frac{\lambda}{8}}}$
=$\lambda' = \frac{h}{\left(\sqrt{\frac{\lambda}{4}} \times \sqrt{2m}\right)}$
= $\lambda' = \frac{2h}{\sqrt{\lambda \times 2m}}$
We can see that the new wavelength λ' is equal to λ multiplied by a constant factor of 2.

Therefore, the correct option is (D) $2λ.$