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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
Two particles move at right angle to each other. Their de Broglie wavelengths are \({\lambda _1}\) and \({\lambda _2}\) respectively. The particles suffere perfectly inelastic collision. The de Broglie wavelength \(\lambda \) , of the final particle, is given by
- A \(\lambda = \sqrt {{\lambda _1}{\lambda _2}} \)
- B \(\lambda = \frac{{{\lambda _1} + {\lambda _2}}}{2}\)
- C \(\frac{2}{\lambda } = \frac{1}{{{\lambda _1}}} + \frac{1}{{{\lambda _2}}}\)
- D \(\frac{1}{{{\lambda ^2}}} = \frac{1}{{\lambda _1^2}} + \frac{1}{{\lambda _2^2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{{{\lambda ^2}}} = \frac{1}{{\lambda _1^2}} + \frac{1}{{\lambda _2^2}}\)
Step-by-step Solution
Detailed explanation
\(\vec{P}_{1}=\frac{h}{\lambda_{1}} \hat{i}\) and \(\vec{P}_{2}=\frac{h}{\lambda_{2}} \hat{j}\) Using momentum conservation \(\vec{P}=\vec{P}_{1}+\vec{P}_{2}=\frac{h}{\lambda_{1}} \hat{i}+\frac{h}{\lambda_{2}} \hat{j}\)…
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