Particle A of mass m_{A} = frac{m}{2} moving | vedclass

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

Question

(A) Applying the law of conservation of linear momentum:$\frac{m}{2} v_{0} + \frac{m}{3} (0) = \frac{m}{2} v_{A} + \frac{m}{3} v_{B}$$\frac{v_{0}}{2}

Vedclass · Accepted Answer

$\Delta \lambda = 4 \lambda_{0}$

Vedclass · Answer

(A) Applying the law of conservation of linear momentum:$\frac{m}{2} v_{0} + \frac{m}{3} (0) = \frac{m}{2} v_{A} + \frac{m}{3} v_{B}$$\frac{v_{0}}{2} = \frac{v_{A}}{2} + \frac{v_{B}}{3} \Rightarrow v_{0} = v_{A} + \frac{2}{3} v_{B} \Rightarrow 3v_{0} = 3v_{A} + 2v_{B} \quad ....(1)$Since the collision is elastic $(e = 1)$:$e = 1 = \frac{v_{B} - v_{A}}{v_{0}} \Rightarrow v_{0} = v_{B} - v_{A} \quad ....(2)$From equation $(2)$,$v_{B} = v_{0} + v_{A}$. Substituting this into equation $(1)$:$3v_{0} = 3v_{A} + 2(v_{0} + v_{A})$$3v_{0} = 3v_{A} + 2v_{0} + 2v_{A}$$v_{0} = 5v_{A} \Rightarrow v_{A} = \frac{v_{0}}{5}$Initial de-Broglie wavelength of particle $A$:$\lambda_{0} = \frac{h}{m_{A} v_{0}} = \frac{h}{(\frac{m}{2}) v_{0}} = \frac{2h}{mv_{0}}$Final de-Broglie wavelength of particle $A$:$\lambda_{f} = \frac{h}{m_{A} v_{A}} = \frac{h}{(\frac{m}{2}) (\frac{v_{0}}{5})} = \frac{10h}{mv_{0}}$Change in de-Broglie wavelength:$\Delta \lambda = \lambda_{f} - \lambda_{0} = \frac{10h}{mv_{0}} - \frac{2h}{mv_{0}} = \frac{8h}{mv_{0}}$Since $\lambda_{0} = \frac{2h}{mv_{0}}$,then $\Delta \lambda = 4 	imes (\frac{2h}{mv_{0}}) = 4 \lambda_{0}$.

Similar Questions

The graph shows the variation of de-Broglie wavelength $\lambda$ versus $\frac{1}{\sqrt{V}}$,where $V$ is the accelerating potential for four particles carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents the particle with the smallest mass?

Proton $(P)$ and electron $(e)$ will have the same de Broglie wavelength when the ratio of their momentum is (assume,$m_{p} = 1849 \, m_{e}$)

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 $..............$.

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

The ratio of the de-Broglie wavelengths of a proton and an electron having the same kinetic energy is: (Assume $m_p = 1849 \times m_e$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module