Particle A of mass


Particle $A$ of mass $m_{A}=\frac{m}{2}$ moving along the $x$-axis with velocity $v_{0}$ collides elastically 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-Broglie wavelength of particle $A$, in terms of its de-Broglie wavelength $\left(\lambda_{0}\right)$ before collision is :

  1. (1) $\Delta \lambda=\frac{3}{2} \lambda_{0}$

  2. (2) $\Delta \lambda=\frac{5}{2} \lambda_{0}$

  3. (3) $\Delta \lambda=2 \lambda_{0}$

  4. (4) $\Delta \lambda=4 \lambda_{0}$

Correct Option: , 4


Applying momentum conservation

$\frac{m}{2} \times V_{0}+\frac{m}{3} \times(0)=\frac{m}{2} V_{A}+\frac{m}{3} V_{B}$

$=\frac{V_{0}}{2}=\frac{V_{A}}{2}+\frac{V_{B}}{3}$         ...(1)

Since, collision is elastic

$e=1=\frac{V_{B}-V_{A}}{V_{0}} \Rightarrow V_{0}=V_{B}-V_{A}$         ...(2)

On solving equations (i) and (ii) $: V_{A}=\frac{V_{0}}{5}$

Now, de-Broglie wavelength of $A$ before collision :

$\lambda_{0}=\frac{h}{m_{A} V_{0}}=\frac{h}{\left(\frac{m}{2}\right) V_{0}} \Rightarrow \lambda_{0}=\frac{2 h}{m V_{0}}$

Final de-Broglie wavelength :

$\lambda_{f}=\frac{h}{m_{A} V_{0}}=\frac{h}{\frac{m}{2} \times \frac{V_{0}}{5}} \Rightarrow \lambda_{f}=\frac{10 h}{m V_{0}}$

$\therefore \Delta \lambda=\lambda_{f}-\lambda_{0}=\frac{10 h}{m V_{0}}-\frac{2 h}{m V_{0}}$

$\Rightarrow \Delta \lambda=\frac{8 h}{m v_{0}} \Rightarrow \Delta \lambda=4 \times \frac{2 h}{m v_{0}}$

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

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