COMEDK · Physics · 27. Dual Nature of Matter

De Broglie wavelength associated with an electron accelerated through a potential difference ' V ' is ' \(\lambda\) '. If the accelerating potential is halved, what will be the new wavelength associated with the charged particle?

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

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

The de Broglie wavelength \(\lambda\) of an electron accelerated through a potential difference \(V\) is given by the formula \(\lambda = \dfrac{h}{\sqrt{2meV}}\), where \(h\) is Planck's constant, \(m\) is the mass of the electron, and \(e\) is the charge of the electron. From…

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.

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