COMEDK · Physics · 27. Dual Nature of Matter

The de-Broglie wavelength of a neutron having kinetic energy 0.025 eV is \(\lambda\). If the kinetic energy of the neutron is reduced to \(\dfrac{0.025 \mathrm{eV}}{8}\), its de-Broglie wavelength will be:

A \(2 \sqrt{2} \lambda\)
B \(\dfrac{\lambda}{2 \sqrt{2}}\)
C \(\dfrac{0.025 \lambda}{2 \sqrt{2}}\)
D \(\dfrac{\lambda}{0.025 \sqrt{2}}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(2 \sqrt{2} \lambda\)

Step-by-step Solution

Detailed explanation

The de-Broglie wavelength \(\lambda\) of a particle with mass \(m\) and kinetic energy \(K\) is given by the relation \(\lambda = \dfrac{h}{\sqrt{2mK}}\). For the initial state, the wavelength is \(\lambda = \dfrac{h}{\sqrt{2m(0.025 \text{ eV})}}\). When the kinetic energy is…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From COMEDK

Explore more questions on app