A particle A of mass m and initial velocity v | vedclass

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

Question

(B) Given: Mass of $A = m$,Mass of $B = \frac{m}{2}$. Initial velocity of $A = v$,Initial velocity of $B = 0$.Let the final velocities be $v_1$ and $v

Vedclass · Accepted Answer

$\frac{\lambda_A}{\lambda_B} = 2$

Vedclass · Answer

(B) Given: Mass of $A = m$,Mass of $B = \frac{m}{2}$. Initial velocity of $A = v$,Initial velocity of $B = 0$.Let the final velocities be $v_1$ and $v_2$ for $A$ and $B$ respectively.By conservation of momentum: $mv = mv_1 + (\frac{m}{2})v_2 \implies v = v_1 + \frac{v_2}{2} \implies 2v = 2v_1 + v_2$ ... $(i)$.For an elastic collision,the coefficient of restitution $e = 1$,so $v_2 - v_1 = v - 0 \implies v_2 = v + v_1$ ... $(ii)$.Substituting $(ii)$ into $(i)$: $2v = 2v_1 + (v + v_1) \implies v = 3v_1 \implies v_1 = \frac{v}{3}$.Then $v_2 = v + \frac{v}{3} = \frac{4v}{3}$.The de-Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$.Ratio $\frac{\lambda_A}{\lambda_B} = \frac{p_B}{p_A} = \frac{m_B v_2}{m_A v_1} = \frac{(\frac{m}{2}) 	imes (\frac{4v}{3})}{m 	imes (\frac{v}{3})} = \frac{\frac{2mv}{3}}{\frac{mv}{3}} = 2$.

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ and $\lambda_B$ after the collision is

Similar Questions

The wavelength of a charged particle of mass $8.0 \times 10^{-31} \ kg$, charge $1.6 \times 10^{-19} \ C$ and kinetic energy $3 \ keV$ will be (Planck constant, $h = 6.4 \times 10^{-34} \ Js$) (in $\text{Å}$)

The interatomic spacing in a crystal is $1.227 \ \mathring A$. What is the maximum order of diffraction for electrons accelerated by $10 \ kV$?

The de Broglie wavelength of a molecule in a gas at room temperature $(300 \ K)$ is $\lambda_1$. If the temperature of the gas is increased to $600 \ K$,then the de Broglie wavelength of the same gas molecule becomes $..............$.

The relation between the wavelength of electromagnetic radiation $(\lambda)$ and the de Broglie wavelength of its quantum (photon) $(\lambda')$ is . . . . . . .

The de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda_0$. What will be its wavelength at $927^{\circ} C$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module