COMEDK · Physics · 27. Dual Nature of Matter
A proton accelerated through a potential \(V\) has de-Broglie wavelength \(\lambda\). Then, the de-Broglie wavelength of an \(\alpha\)-particle, when accelerated through the same potential \(V\) is
- A \(\frac{\lambda}{2}\)
- B \(\frac{\lambda}{\sqrt{2}}\)
- C \(\frac{\lambda}{2 \sqrt{2}}\)
- D \(\frac{\lambda}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{\lambda}{2 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
When a charged particle of charge \(q\) is accelerated through a potential \(V\), then de-Broglie wavelength is given as \(\lambda=\frac{h}{\sqrt{2 m q V}}\) \(\Rightarrow \quad \lambda \propto \frac{1}{\sqrt{m q}}\)…
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