Question

If de-Broglie wavelength is $\lambda$ when energy is E. Find the wavelength at $\frac{E}{4}$ (Kinetic Energy).

• A. $2\lambda$
• B. $\sqrt{2}\lambda$
• C. $\lambda$
• D. $\frac{\lambda }{\sqrt{2}}$

Answer 1

Answer: (A)

$2\lambda$

Answer 2

The de Broglie wavelength of a particle is given by the formula:
$\lambda=\frac{h}{p}$
where h is Planck's constant and p is the momentum of the particle. The momentum of a particle with kinetic energy E is given by:
$p = √(2mE)$
where m is the mass of the particle.
If the de Broglie wavelength of a particle with energy E is 𝜆, then we have:
$\lambda = \frac{h}{√(2mE)}$
To find the de Broglie wavelength of a particle with kinetic energy E/4, we first need to find the momentum of the particle. The momentum is given by:
$p = \sqrt{(2m(\frac{E}{4}))} = \sqrt{(m\frac{E}{2})}$
The de Broglie wavelength of the particle with momentum p is then given by:
$\lambda^{'} = \frac{h}{p} = \frac{h}{\sqrt(m\frac{E}{2})}$
Dividing this expression by the original expression for 𝜆, we get:
$\frac{{\lambda}'}{\lambda} = \sqrt{2}$
So, the wavelength of the particle with kinetic energy $\frac{E}{4}$ is √2 times the wavelength of the particle with kinetic energy E. Therefore, the answer is option 2: √2𝜆.
**Answer. **A