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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are \({\lambda _1}\) and \({\lambda _2}\), their de Broglie wavelength in the frame of reference attached to their centre of mass is
- A \({\lambda _{CM}} = {\lambda _1} = {\lambda _2}\)
- B \(\frac{1}{{{\lambda _1}}} = \frac{1}{{{\lambda _1}}} + \frac{1}{{{\lambda _2}}}\)
- C \({\lambda _{CM}} = \frac{{2{\lambda _1}{\lambda _2}}}{{\sqrt {\lambda _1^2 + \lambda _2^2} }}\)
- D \({\lambda _{CM}} = \left( {\frac{{{\lambda _1} + {\lambda _2}}}{2}} \right)\)
Answer & Solution
Correct Answer
(C) \({\lambda _{CM}} = \frac{{2{\lambda _1}{\lambda _2}}}{{\sqrt {\lambda _1^2 + \lambda _2^2} }}\)
Step-by-step Solution
Detailed explanation
Momentum \((p)\) of each electron \(\frac{\mathrm{h}}{\lambda_{1}} \hat{\mathrm{i}}\) and \(\frac{\mathrm{h}}{\lambda_{2}} \hat{\mathrm{j}}\) Velocity of centre of mass \({{\text{V}}_{{\text{cm}}}} = \frac{{\text{h}}}{{2{\text{m}}{\lambda _1}}}\widehat {\text{i}} + \)…
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