Two particles move at right angle to each other. Their de Broglie wavelengths are λ 1 and λ 2 respectively. The particles suffere perfectly inelastic collision. The de Broglie wavelength λ, 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}}}

JEE Mains · Physics · STD 12 - 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

Pick an option to check your answer.

Verified Solution

Correct Answer

(D) \(\frac{1}{{{\lambda ^2}}} = \frac{1}{{\lambda _1^2}} + \frac{1}{{\lambda _2^2}}\)

Detailed explanation

\(\vec{P}_{1}=\frac{h}{\lambda_{1}} \hat{i}\) and \(\vec{P}_{2}=\frac{h}{\lambda_{2}} \hat{j}\)

Using momentum conservation

\(\vec{P}=\vec{P}_{1}+\vec{P}_{2}=\frac{h}{\lambda_{1}} \hat{i}+\frac{h}{\lambda_{2}} \hat{j}\)

\(|\vec P| = \sqrt {{{\left( {\frac{h}{{{\lambda _1}}}} \right)}^2} + {{\left( {\frac{h}{{{\lambda _2}}}} \right)}^2}} \)

\(\frac{h}{\lambda}=\sqrt{\left(\frac{h}{\lambda_{1}}\right)^{2}+\left(\frac{h}{\lambda_{2}}\right)^{2}}\)

\(\frac{1}{\lambda^{2}}=\frac{1}{\lambda_{1}^{2}}+\frac{1}{\lambda_{2}^{2}}\)

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