A particle P is formed due to a completely in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to the principle of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collisio

Vedclass · Accepted Answer

$\frac{\lambda_x \lambda_y}{|\lambda_x - \lambda_y|}$

Vedclass · Answer

(C) According to the principle of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collision.Let the momenta of particles $x$ and $y$ be $\vec{p}_x$ and $\vec{p}_y$ respectively.Since they move in opposite directions,we take one direction as positive and the other as negative.The magnitude of momentum is related to the de-Broglie wavelength by $p = \frac{h}{\lambda}$.Let $p_x = \frac{h}{\lambda_x}$ and $p_y = \frac{h}{\lambda_y}$.The final momentum of particle $P$ is $p_f = |p_x - p_y|$.Substituting the values,$p_f = |\frac{h}{\lambda_x} - \frac{h}{\lambda_y}| = h |\frac{1}{\lambda_x} - \frac{1}{\lambda_y}|$.Since $p_f = \frac{h}{\lambda}$,we have $\frac{h}{\lambda} = h |\frac{\lambda_y - \lambda_x}{\lambda_x \lambda_y}|$.Therefore,$\lambda = \frac{\lambda_x \lambda_y}{|\lambda_x - \lambda_y|}$.

$A$ particle $P$ is formed due to a completely inelastic collision of particles $x$ 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

Similar Questions

An electron is moving with an initial velocity $\vec{V} = V_{0} \hat{i}$ and is in a uniform magnetic field $\vec{B} = B_{0} \hat{j}$. Then its de Broglie wavelength

Which one of the following statements is not true about de-Broglie waves?

The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)

Compute the typical de Broglie wavelength of an electron in a metal at $27\,^{\circ} C$ and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{-10} \; m$.

If $\lambda$ and $K$ are the de Broglie wavelength and kinetic energy,respectively,of a particle with constant mass,the correct graphical representation for the particle will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module