First slide

Wave nature of matter

Question

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 the particles move along the x-axis after the collision, the change $Δ λ$ in de-Broglie wavelength of particle A, in terms of its de-Broglie wavelength $(λ_{0})$ before collision is :

Moderate

A

$Δ λ = 2 λ_{0}$
B

$Δ λ = \frac{5}{2} λ_{0}$
C

$Δ λ = 4 λ_{0}$
D

$Δ λ = \frac{3}{2} λ_{0}$

Solution

$\begin{array}{l} C o n s e r v a t i o n o f m o m e n t u m, \\ \frac{m}{2} V_{0} + \frac{m}{3} (0) = \frac{m}{2} (V_{A}) + \frac{m}{3} (V_{B}) \\ \Rightarrow 3 V_{A} + 2 V_{B} = 3 V_{0} ..................... (1) \\ C o e f f i c i e n t o f r e s t i t u t i o n , \\ e = 1 = \frac{V_{A} - V_{B}}{0 - V_{0}} \\ \Rightarrow V_{A} - V_{B} = - V_{0} ....................... (2) \\ S o l v i n g (1) a n d (2) \\ V_{A} = \frac{V_{0}}{5} \\ N o w, d e b r o g l i e w a v e l e n g t h = λ = \frac{h}{m v} \\ S o, I n i t i a l w a v e l e n g t h o f A = λ_{i} = \frac{h}{\frac{m}{2} V_{0}} = λ_{0} \\ a n d F i n a l w a v e l e n g t h o f A = λ_{f} = \frac{h}{\frac{m}{2} \frac{V_{0}}{5}} = 5 λ_{0} \\ T h u s, C h a n g e i n w a v e l e n g t h = △ λ = λ_{f} - λ_{i} = 4 λ_{0} \end{array}$

In ground frame, $\frac{m}{2}$ moves forward with $\frac{3 V_{0}}{5} - \frac{2 V_{0}}{5} = \frac{V_{0}}{5}$
So $λ_{f} = 5 λ_{0} \Rightarrow Δ λ = 4 λ_{0}$

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