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  • Two particles A and B of de Broglie wavelengths λ1 and λ2 combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).

Two particles A and B of de Broglie wavelengths \lambda_{1} and \lambda_{2} combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).

Answers (1)

The de-Broglie wavelength is given by

\lambda=\frac{h}{p} \; or\; p=\frac{h}{\lambda}

p_{1}=\frac{h}{\lambda_{1}}, p_{2}=\frac{h}{\lambda_{2}}, p_{3}=\frac{h}{\lambda_{3}}
By the law of conservation of momentum
p_{1}+p_{2}= p_{3}

\frac{h}{\lambda_{1}}+\frac{h}{\lambda_{2}}=\frac{h}{\lambda_{3}}  (\lambda_{3} is the wavelength of particle C)
 

\frac{h}{\lambda_{1}}+\frac{h}{\lambda_{2}}=\frac{h}{\lambda_{3}}\\ \frac{\lambda_{2}+\lambda_{1}}{\lambda_{1}\lambda_{2}}=\frac{h}{\lambda_{3}} 

Case I:When p1 and p2 are positive then \lambda_{3}=\frac{\lambda_{1}\lambda_{2}}{\lambda_{1}+\lambda_{2}}

  Case II:When p1 and p2 both are negative \lambda_{3}=\left |-\frac{\lambda_{1}\lambda_{2}}{\lambda_{1}+\lambda_{2}} \right |=\frac{\lambda_{1}\lambda_{2}}{\lambda_{1}+\lambda_{2}}

 Case III: p_{A}>0,p_{B} <0

\frac{h}{\lambda_{3}}=\frac{h}{\lambda_{1}}-\frac{h}{\lambda_{2}}\; or\; \frac{h}{\lambda_{3}}=\frac{(\lambda_{2}-\lambda_{1})h}{\lambda_{1}\lambda_{2}}

Case IV: p_{A}<0,p_{B} >0

\frac{h}{\lambda_{3}}=\frac{h}{\lambda_{1}}-\frac{h}{\lambda_{2}}=\frac{\left (\lambda_{1}-\lambda_{2} \right )h}{\lambda_{1}\lambda_{2}}\\ \lambda_{3}=\frac{\lambda_{1}\lambda_{2}}{\lambda_{1}-\lambda_{2} }

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infoexpert24

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