TS EAMCET · Physics · Dual Nature of Matter
A particle of charge \(q\), mass \(m\) and energy \(E\) has de-Broglie wavelength \(\lambda\). For a particle of charge \(2 q\), mass \(2 m\) and energy \(2 E\), the de-Broglie wavelength is
- A \(\frac{\lambda}{4}\)
- B \(2 \lambda\)
- C \(8 \lambda\)
- D \(\frac{\lambda}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\lambda}{2}\)
Step-by-step Solution
Detailed explanation
The de-Broglie wavelength \(\lambda\) of charged particle of charge \(q\) mass \(m\) and energy \(E\) is given by \[ \lambda=\frac{h}{\sqrt{2 m E}} \] For another particle mass \(m^{\prime}=2 \mathrm{~m}\) charge, \(q^{\prime}=2 q\) and energy, \(B^{\prime}=2 B\) de-Broglie…
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