Question

A particle $'P'$ is formed due to a completely inelastic collision of particle $'x'$ and $'y'$ having de-Broglie wavelengths $'{{\lambda }_{x}}'$ and $'{{\lambda }_{y}}'$ respectively. If $x$ and $y$ were moving in opposite directions, then the de-Broglie wavelength of $'P'$ is?
$A.{{\lambda }_{x}}+{{\lambda }_{y}}$
$B.\dfrac{{{\lambda }_{x}}{{\lambda }_{y}}}{{{\lambda }_{x}}+{{\lambda }_{y}}}$
$C.\dfrac{{{\lambda }_{x}}{{\lambda }_{y}}}{\left| {{\lambda }_{x}}-{{\lambda }_{y}} \right|}$
$D.{{\lambda }_{x}}-{{\lambda }_{y}}$

Answer 1

Hint: We know that the principle of conservation of momentum states that for an isolated system momentum can neither be created nor be destroyed. It means that total momentum before collision is equal to total momentum after collision. As the collision is inelastic the kinetic energy is not conserved.
Formula used:
 This problem can be solved by using the following formula:-
$\lambda =\dfrac{h}{p}$.

Complete answer:
Using de-Broglie equation for a particle which is also considered as wave,
$\lambda =\dfrac{h}{p}$…………….. $(i)$
From this equation we get,
$p=\dfrac{h}{\lambda }$Where, $p$is the momentum of the particle, $h$ is Planck's constant and $\lambda $is wavelength of the particle.
For particle $'x'$we have,
${{p}_{x}}=\dfrac{h}{{{\lambda }_{x}}}$………………… $(ii)$
For particle $'y'$we have,
${{p}_{y}}=\dfrac{h}{{{\lambda }_{y}}}$……………….. $(iii)$
From the principle of conservation of momentum we have,
${{p}_{x}}+{{p}_{y}}=p$……………… $(iv)$
Using equation $(ii)$and $(iii)$we can write the equation $(iv)$as follows:-
$\dfrac{h}{{{\lambda }_{x}}}-\dfrac{h}{{{\lambda }_{y}}}=\dfrac{h}{\lambda }$(Negative sign is for fact that the both the particles are moving in opposite direction). $\lambda $ represents the total wavelength of the particles after the collision.
Solving further, we get
$\dfrac{{{\lambda }_{y}}h-{{\lambda }_{x}}h}{{{\lambda }_{x}}{{\lambda }_{y}}}=\dfrac{h}{\lambda }$
$\dfrac{h({{\lambda }_{y}}-{{\lambda }_{x}})}{{{\lambda }_{x}}{{\lambda }_{y}}}=\dfrac{h}{\lambda }$
Simplifying we get,
$\dfrac{({{\lambda }_{y}}-{{\lambda }_{x}})}{{{\lambda }_{x}}{{\lambda }_{y}}}=\dfrac{1}{\lambda }$
Therefore, $\lambda =\dfrac{{{\lambda }_{x}}{{\lambda }_{y}}}{{{\lambda }_{y}}-{{\lambda }_{x}}}$

So, the correct answer is “Option C”.

Additional Information:
Collision is defined as the sudden coming together of two objects for a short period of time. There are two types of collisions, one is elastic collision and other is inelastic collision. Elastic collision refers to the situation where kinetic energy remains conserved during the process while inelastic collision is the situation where kinetic energy does not remain conserved.

Note:
In solving these types of problems we have to take care about the fact that whether the collision is elastic or inelastic. Directions of colliding particles should also be taken care of. This problem is solved by assuming the dual nature of particles using the de-Broglie equation.