Question

A particle A of mass m and initial velocity v collides with a particle B of mass $\dfrac{m}{2}$ which is at rest. The collision is head on and elastic. The ratio of the de-Broglie wavelengths ${{\lambda }_{A}}$ to ${{\lambda }_{B}}$ after the collision is:
$\text{A}\text{. }\dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\dfrac{1}{2}$
$\text{B}\text{. }\dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\dfrac{1}{3}$
$\text{C}\text{. }\dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=2$
$\text{D}\text{. }\dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\dfrac{2}{3}$

Answer 1

Use the law of conservation of momentum. Equate the momentums and kinetic energies of the system before and after the collision and find the velocities of the particles after the collision. Then use the formula for de-Broglie wavelength ($\lambda =\dfrac{h}{mv}$) and find the ratio of the wavelengths.

Formula used:
p=mv
$\lambda =\dfrac{h}{mv}$

Complete answer:
It is given that particle A collides with particle B. It is also given that the collision is head on (in straight line) and elastic. Let assume that there is no force other than the reaction forces when the particle collides, affecting the system of the two particles.
Therefore, apply the law of conservation of momentum. This means that the initial momentum (before collision) and the final momentum (after collision) of the system are equal.
Momentum of a particle is given as the product of its mass and its velocity.
The masses of particle A and particle B are given to be m and $\dfrac{m}{2}$ respectively.

Before the collision, the velocity of particle A is given to be v and particle B is said to be at rest.
Hence, the initial momentum of the system is ${{p}_{i}}=mv$.
After the collision, let the velocity of particle A be ${{v}_{A}}$ and the velocity of particle B be ${{v}_{B}}$.

Hence, the final momentum of the system is ${{p}_{f}}=m{{v}_{A}}+\dfrac{m}{2}{{v}_{B}}$.
And ${{p}_{i}}={{p}_{f}}$.
Therefore, $mv=m{{v}_{A}}+\dfrac{m}{2}{{v}_{B}}$
$\Rightarrow v-\dfrac{{{v}_{B}}}{2}={{v}_{A}}$ …. (i).
Since the collision is elastic, the kinetic energy of the system will be conserved. This means that the kinetic energy of the system before the collision is equal to kinetic energy of the system after the collision.
Before the collision, the kinetic energy of the system is ${{K}_{i}}=\dfrac{1}{2}m{{v}^{2}}$.
After the collision, the kinetic energy of the system is ${{K}_{f}}=\dfrac{1}{2}mv_{A}^{2}+\dfrac{1}{2}\left( \dfrac{m}{2} \right)v_{B}^{2}$.
And ${{K}_{i}}={{K}_{f}}$.

Therefore,
$\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}mv_{A}^{2}+\dfrac{1}{2}\left( \dfrac{m}{2} \right)v_{B}^{2}$
$\Rightarrow 2{{v}^{2}}=2v_{A}^{2}+v_{B}^{2}$ …. (ii).

Substitute the value of ${{v}_{A}}$ from equation (i).
$\Rightarrow 2{{v}^{2}}=2{{\left( v-\dfrac{{{v}_{B}}}{2} \right)}^{2}}+v_{B}^{2}$
$\Rightarrow 2{{v}^{2}}=2\left( {{v}^{2}}-v{{v}_{B}}+\dfrac{v_{B}^{2}}{4} \right)+v_{B}^{2}$
$\Rightarrow 2{{v}^{2}}=2{{v}^{2}}-2v{{v}_{B}}+\dfrac{v_{B}^{2}}{2}+v_{B}^{2}$
$\Rightarrow \dfrac{3v_{B}^{2}}{2}-2v{{v}_{B}}=0$.
$\Rightarrow {{v}_{B}}\left( \dfrac{3{{v}_{B}}}{2}-2v \right)=0$

Therefore, ${{v}_{B}}$= 0 or $\left( \dfrac{3{{v}_{B}}}{2}-2v \right)$= 0.
${{v}_{B}}$ cannot be equal to zero. Hence, ${{v}_{B}}$= 0 is discarded.
Therefore, $\left( \dfrac{3{{v}_{B}}}{2}-2v \right)$= 0.
$\Rightarrow {{v}_{B}}=\dfrac{4v}{3}$.

Substitute value of ${{v}_{B}}$ in equation (i).
$\Rightarrow v-\dfrac{\dfrac{4v}{3}}{2}={{v}_{A}}$
$\Rightarrow v-\dfrac{2v}{3}={{v}_{A}}$
$\Rightarrow {{v}_{A}}=\dfrac{v}{3}$.

The De-Broglie wavelength of a particle is given as $\lambda =\dfrac{h}{mv}$, h is Planck's constant, m and v are its mass and its velocity.

Therefore, ${{\lambda }_{A}}=\dfrac{h}{m{{v}_{A}}}=\dfrac{3h}{mv}$.
${{\lambda }_{B}}=\dfrac{h}{m{{v}_{B}}}=\dfrac{h}{\dfrac{m}{2}.\dfrac{4v}{3}}=\dfrac{3h}{2mv}$.
Therefore, $\Rightarrow \dfrac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\dfrac{\dfrac{3h}{mv}}{\dfrac{3h}{2mv}}=2$.

So, the correct answer is “Option C”.

Note:
According to de-Broglie’s theory, everybody in the universe shows wave nature. However, we do not see any wave form of the bodies in our daily lives. This is because the momentum of large bodies is very large when compared to the value of Planck's constant. Therefore, the wavelengths of the bodies are very small and negligible.