TS EAMCET · Physics · Dual Nature of Matter
A proton when accelerated through a potential difference of \(V\), has a de-Broglie wavelength \(\lambda\) associated with it. If an \(\alpha\)-particle is to have the same de-Broglie wavelength \(\lambda\), it must be accelerated through a potential difference of
- A \(\frac{V}{8}\)
- B \(\frac{V}{4}\)
- C \(4 \mathrm{~V}\)
- D \(8 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(\frac{V}{8}\)
Step-by-step Solution
Detailed explanation
\(\lambda_p=\lambda_\alpha\) \((m q V)_p=(m q V)_\alpha\) Potential difference \(V_\alpha=\frac{V}{8} \quad\left[\begin{array}{c}\because m_\alpha=4 m_p \\ q_\alpha=2 q_p\end{array}\right]\)
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