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JEE Mains · Physics · STD 12 - 12. atoms

A particle \(P\) is formed due to a completely inelastic collision of particles \(x\) and \(y\) having de -Broglie wavelengths \({\lambda _x}\) and \({\lambda _y}\) respectively. If \(x\) and \(y\) were moving in opposite directions, then the de -Broglie wavelength of \(P\) is

  1. A \({\lambda _x} + {\lambda _y}\)
  2. B \(\frac{{{\lambda _x}{\lambda _y}}}{{{\lambda _x} + {\lambda _y}}}\)
  3. C \(\frac{{{\lambda _x}{\lambda _y}}}{{\left| {{\lambda _x} - {\lambda _y}} \right|}}\)
  4. D \({\lambda _x} - {\lambda _y}\)
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Answer & Solution

Correct Answer

(C) \(\frac{{{\lambda _x}{\lambda _y}}}{{\left| {{\lambda _x} - {\lambda _y}} \right|}}\)

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Detailed explanation

Conservation of momentum \({\overrightarrow {\text{p}} _x} + {\overrightarrow {\text{p}} _y} = {\overrightarrow {\text{p}} _{{\text{final}}}}\)…
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