Particle A of mass m_{A}=\frac{m}{2} moving along the axis with velocity v_{0} collides eleastically with another particle B at rest having mass m_{B}=\frac{m}{3}. If both particles move along the x-axis after the collision, the change \Delta \lambda in de-Brogile wavelength of particle A, in terms of its de-Brogile wavelength (\lambda _{0}) before collision is:
Option: 1 \Delta \lambda =\frac{3}{2}\lambda _{0}
Option: 2 \Delta \lambda =\frac{5}{2}\lambda _{0}
Option: 3 \Delta \lambda =2\lambda _{0}
Option: 4 \Delta \lambda =4\lambda _{0}

Answers (1)

Applying momentum conservation
\frac{\mathrm{m}}{2} \times \mathrm{V}_{0}+\frac{\mathrm{m}}{3} \times(0)=\frac{\mathrm{m}}{2} \mathrm{~V}_{\mathrm{A}}+\frac{\mathrm{m}}{3} \mathrm{~V}_{\mathrm{B}}$ \\ $\Rightarrow \frac{\mathrm{V}_{0}}{2}=\frac{\mathrm{V}_{\mathrm{A}}}{2}+\frac{\mathrm{V}_{\mathrm{B}}}{3} \quad \ldots (1)
Since, collision is elastic (e=1)
\mathrm{e}=1=\frac{\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{0}} \\ \Rightarrow \mathrm{V}_{0}=\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}} \ldots (2)

From equation (1) and (2)

we get

V_A=\frac{V_0}{5}

Now, De-Broglie wavelength of A before collision: 

\begin{array}{l} \lambda_{0}=\frac{h}{m_{A} V_{0}}=\frac{h}{\left(\frac{m}{2}\right) V_{0}} \\ \\ \Rightarrow \lambda_{0}=\frac{2 h}{m V_{0}} \end{array}

Final De-Broglie wavelength
\lambda_{\mathrm{f}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{V}_{0}}=\frac{\mathrm{h}}{\frac{\mathrm{m}}{2} \times \frac{\mathrm{V}_{0}}{5}} \\ \\ \Rightarrow \lambda_{\mathrm{f}}=\frac{10 \mathrm{~h}}{\mathrm{mV}_{0}}

\begin{array}{l} \text { Now } \Delta \lambda=\lambda_{\mathrm{f}}-\lambda_{0} \\ \Delta \lambda=\frac{10 \mathrm{~h}}{\mathrm{mV}_{0}}-\frac{2 \mathrm{~h}}{\mathrm{mV}_{0}} \\ \Rightarrow \Delta \lambda=\frac{8 \mathrm{~h}}{\mathrm{mv}_{0}} \\\Rightarrow \Delta \lambda=4 \times \frac{2 \mathrm{~h}}{\mathrm{mv}_{0}} \\ \Rightarrow \Delta \lambda=4 \lambda_{0} \end{array}

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