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  • A particle  is formed due to a completely inelastic collision of particles <img alt="'x'and\, 'y'"

A particle 'p' is formed due to a completely inelastic collision of particles 'x'and\, 'y' having de-broglie wavelength '\lambda _{x}'and\, '\lambda _{y}' respectively . If x\: and\; y were moving in opposite directions , then the de-broglie wavelength of 'p' is :

Option: 1

\frac{\lambda _{x}\lambda _{y}}{\lambda _{x}+\lambda _{y}}


Option: 2

\frac{\lambda _{x}\lambda _{y}}{\left | \lambda _{x}-\lambda _{y} \right |}


Option: 3

\lambda _{x}-\lambda _{y}


Option: 4

\lambda _{x}+\lambda _{y}


Answers (1)

best_answer

Using de-Broglie equation for a particle which is also considered as wave,
\lambda= \frac{h}{p} \quad \dots \dots\left ( i \right )
From this equation, we get
p= \frac{h}{\lambda}
where p is the momentum of the particle, h is Planck's constant, and\lambda: is the wavelength of the particle.
For particle x, we have
p_{x}= \frac{h}{\lambda_{x}} \dots\dots\left ( ii \right )
For particle y , we have
p_{y}= \frac{h}{\lambda_{y}} \dots\dots\left ( iii \right )
From the principle of conservation of momentum, we have
p_{x}+p_{y}= p\dots\dots\left ( iv\right )

Using equations (ii) and (iii):
\frac{h}{\lambda_{x}}-\frac{h}{\lambda_{y}}= \frac{h}{\lambda}

(Negative sign is for fact that the both the particles are moving in opposite direction)

Here, \lambda: represents the total wavelength of the particles after the collision.
Solving further, we get
\frac{\lambda_{y}h-\lambda_{x}h}{\lambda_{x}\lambda_{y}}= \frac{h}{\lambda}
\frac{\lambda_{y}-\lambda_{x}}{\lambda_{x}\lambda_{y}}= \frac{1}{\lambda}
\lambda= \frac{\lambda_{x}\lambda_{y}}{\lambda_{y}-\lambda_{x}}

Posted by

Gaurav

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