A particle P is formed due to a completely inelastic collision of particles x and y having de -Broglie wavelengths λ x and λ y respectively. If x and y were moving in opposite directions, then the de -Broglie wavelength of P is

(A) {\lambda _x} + {\lambda _y}

(B) \frac{{{\lambda _x}{\lambda _y}}}{{{\lambda _x} + {\lambda _y}}}

(D) {\lambda _x} - {\lambda _y}

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JEE Mains · Physics · STD 12 - Atoms

JEE Mains 2019
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MCQ

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

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Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Conservation of momentum

\({\overrightarrow {\text{p}} _x} + {\overrightarrow {\text{p}} _y} = {\overrightarrow {\text{p}} _{{\text{final}}}}\)

\(\mathrm{m}_{\mathrm{x}} \mathrm{V}_{\mathrm{x}}-\mathrm{m}_{\mathrm{y}} \mathrm{V}_{\mathrm{y}}=\left(\mathrm{m}_{\mathrm{x}}+\mathrm{m}_{\mathrm{y}}\right) \mathrm{V}\)

\(\frac{\mathrm{h}}{\lambda_{\mathrm{x}}}-\frac{\mathrm{h}}{\lambda_{\mathrm{y}}}=\frac{\mathrm{h}}{\lambda}\)

\(\Rightarrow \lambda=\frac{\lambda_{x} \lambda_{y}}{\left|\lambda_{x}-\lambda_{y}\right|}\)

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