Question

Choose the correct answer from the alternatives given.
Two particles A and B of de Broglie wavelengths ${\lambda }_A$ and ${\lambda }_B$ combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).

• A. ${\lambda }_A$
• B. ${\lambda }_A{\lambda }_B/({\lambda }_A+{\lambda }_B)$
• C. ${\lambda }_A{\lambda }_B/|{\lambda }_A-{\lambda }_B|$
• D. both (b) and (c)

Answer 1

The correct option is D both (b) and (c)

For one dimensional motion,
$\overrightarrow{PC}=\overrightarrow{P_A}+\overrightarrow{P_B}$
$If{P}_A,P_B>0or{P}_A,P_B<0,i.e.,P_{A}an{P}_{B}areinsamedirection.$

$P_C=P_A+P_B$

$\frac{h}{{\lambda }_C}=\frac{h}{{\lambda }_A}+\frac{h}{{\lambda }_B}=h\left(\frac{{\lambda }_A+{\lambda }_B}{{\lambda }_A{\lambda }_B}\right)\Rightarrow {\lambda }_C=\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

If $P_A>0,P_B<0or{P}_A<0,P_B>0$
$\left(P_{A}and{P}_{B}areinoppositedirection\right)$

$P_C=|P_A-P_B|$

$\frac{h}{{\lambda }_C}=|\frac{h}{{\lambda }_A}-\frac{h}{{\lambda }_B}|=\frac{h|{\lambda }_A-{\lambda }_B|}{{\lambda }_A{\lambda }_B}$

${\lambda }_C=\frac{{\lambda }_A{\lambda }_B}{|{\lambda }_A-{\lambda }_B|}$