TS EAMCET · Physics · Dual Nature of Matter
If \(\lambda_0\) is the de-Broglie wavelength for a proton accelerated through a potential difference of \(100 \mathrm{~V}\), the de-Broglie wavelength for \(\alpha\)-particle accelerated through the same potential difference is
- A \(2 \sqrt{2 \lambda_0}\)
- B \(\frac{\lambda_0}{2}\)
- C \(\frac{\lambda_0}{2 \sqrt{2}}\)
- D \(\frac{\lambda_0}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\lambda_0}{2 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\[ V=100 \text { volt } \] Given, \(\lambda_0=\) de-Broglie wavelength of proton \[ \therefore \quad \lambda_0=\frac{h}{\sqrt{2 m_p q_p V}} \] de-Broglie wavelength of \(\alpha\)-particle…
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