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Two identical particles A & B move at right angle to each other, processing de-broglie wavelength \lambda_1 and \lambda_2, respectively. The de-broglie wavelength of each particle in their centre of mass frame of reference is

Option: 1

\lambda_1+\lambda_2


Option: 2

\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1{ }^2+\lambda_2{ }^2}}


Option: 3

\frac{\lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}


Option: 4

\frac{\lambda_1+\lambda_2}{2}


Answers (1)

best_answer

Let  m  be the mass of each particle then,

\mathrm{\lambda_1=\frac{h}{m v_1}, \lambda_2=\frac{h}{m v_2}}

\mathrm{\vec{v}_{c m}=\frac{m \vec{v}_1+m \vec{v}_2}{2 m}=\frac{\overrightarrow{v_1}+\overrightarrow{v_2}}{2} .}

velocity of  A  w.r.t. Centre frame is 

\mathrm{\begin{aligned} \vec{V}_{1 c} & =\overrightarrow{v_1}-\overrightarrow{v_c}=\frac{\overrightarrow{v_1}-\overrightarrow{v_2}}{2}=\vec{v}_{2 c} \\ \left|\overrightarrow{v_{1 c}}\right| & =\left|\overrightarrow{v_{2 c}}\right|=\sqrt{u_1{ }^2+v_2{ }^2}= \\ & =\frac{h}{m} \cdot \sqrt{\frac{1}{\lambda_1{ }^2}+\frac{1}{\lambda_2{ }^2}} . \end{aligned}}

\mathrm{\lambda=\frac{2 h}{\left.m \mid \overrightarrow{v_1 c}\right)}=\frac{h}{m} \cdot \frac{2}{\frac{h}{m} \sqrt{\frac{1}{\lambda_1{ }^2}+\frac{1}{\lambda_2{ }^2}}} \Rightarrow \lambda=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2{ }^2}} .}

Posted by

rishi.raj

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