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

The Correct Option is A

To solve this problem, we need to understand the relationship between the de Broglie wavelength \(\lambda\) and the kinetic energy \(E\) of a particle.

The de Broglie wavelength \(\lambda\) 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 \(p\) can also be expressed in terms of kinetic energy \(E\) as:

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

where \(m\) is the mass of the particle.

Substituting the expression for momentum into the de Broglie wavelength formula, we get:

\(\lambda = \frac{h}{\sqrt{2mE}}\)

Now, we need to find what the wavelength becomes when the kinetic energy is \(\frac{E}{4}\).

1. Substitute \(\frac{E}{4}\) for \(E\) in the wavelength equation:
2. Simplify the denominator:
3. Further simplify the expression:
4. Recognize that \(\sqrt{2mE} = \frac{h}{\lambda}\) from the original condition:
5. Since \(\frac{h}{\sqrt{2mE}} = \lambda\), we find:

However, there seems to be an oversight in this step. Let's re-evaluate from the point we substituted kinetic energy:

- Let's re-calculate when substituting from: \(\lambda' = 2 \cdot \lambda\)

Thus, the correct wavelength \(\lambda'\) when the energy is reduced to \(\frac{E}{4}\) is:

\(\lambda' = 2\lambda\)

Therefore, the correct answer is indeed:

\(2\lambda\)