- JEE Mains 2018
- Hard
- MCQ
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
Answer & Solution
(C) \({\lambda _{CM}} = \frac{{2{\lambda _1}{\lambda _2}}}{{\sqrt {\lambda _1^2 + \lambda _2^2} }}\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{h}}{\lambda_{1}} \hat{\mathrm{i}}\) and \(\frac{\mathrm{h}}{\lambda_{2}} \hat{\mathrm{j}}\)
Velocity of centre of mass
\({{\text{V}}_{{\text{cm}}}} = \frac{{\text{h}}}{{2{\text{m}}{\lambda _1}}}\widehat {\text{i}} + \) \(\frac{{\text{h}}}{{2{\text{m}}{\lambda _2}}}\widehat {\text{j}}\) \((\because {\text{p}} = {\text{mv}})\)
Velocity of Ist particleabout centre of mass
\(\mathrm{V}_{1 \mathrm{cm}}=\frac{\mathrm{h}}{2 \mathrm{m} \lambda_{1}} \hat{\mathrm{i}}-\frac{\mathrm{h}}{2 \mathrm{m} \lambda_{2}} \hat{\mathrm{j}}\)
\({\lambda _{{\text{cm}}}} = \frac{{\text{h}}}{{\sqrt {\frac{{{{\text{h}}^2}}}{{4\lambda _1^2}} + \frac{{{{\text{h}}^2}}}{{4\lambda _2^2}}} }}\) \( = - \frac{{2{\lambda _1}{\lambda _2}}}{{\sqrt {\lambda _1^2 + \lambda _2^2} }}\left( {\because \lambda = \frac{{\text{h}}}{{\text{p}}}} \right)\)
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