Dual Nature Of Light Question 413
Question: Two identical non-relative particles A and B move right angles to each other, processing de Broglie wavelength $ {\lambda_{1}} $ and $ ,{\lambda_{2}} $ , respectively. The de Broglie wavelength of each particle in their center of mass frame of reference is
Options:
A) $ {\lambda_{1}}+{\lambda_{2}} $
B) $ 2{\lambda_{1}}{\lambda_{2}}/( \sqrt{\lambda _{1}^{2}+\lambda _{2}^{2}} ) $
C) $ 2{\lambda_{1}}{\lambda_{2}}/( \sqrt{| \lambda _{1}^{2}+\lambda _{2}^{2} |} ) $
D) $ ( {\lambda_{1}}+{\lambda_{2}} )/2 $
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Answer:
Correct Answer: B
Solution:
[b] Let m be the mass of each particle, then $ {\lambda_{1}}=( h/mv_{1} ) $ and $ {\lambda_{2}}=( h/mv_{2} ) $ where $ v_{1} $ and $ v_{2} $ are the velocities of two particles as shown in the figure.
$ {\vec{v}_{CM}}=\frac{m{\vec{v}_1}+m{\vec{v}_2}}{2m}$=$\frac{{{{\vec{v}}}_1}-{{{\vec{v}}}_2}}{2} $
$ {\vec{v}_{1c}}$=${\vec{v}1}$-${\vec{v}{CM}}$
=$\frac{{{{\vec{v}}}_1}-{{{\vec{v}}}_2}}{2} $
Velocity of A w.r.t. C frame is $ {{\vec{v}}{1c}}={{\vec{v}}{1}}-{{\vec{v}}{CM}}=\frac{{{{\vec{v}}}{1}}-{{{\vec{v}}}_{2}}}{2} $
$ | {{{\vec{v}}}{1c}} |=\frac{\sqrt{{{{\vec{v}}}{1}}-{{{\vec{v}}}{2}}}}{2}=,| {{{\vec{v}}}{2c}} | $
So, required wavelength is $ \lambda =\frac{h}{3|{{{\vec{v}}}_{1c}}|}=\frac{h}{m}\times \frac{2}{\frac{h}{m}\sqrt{\frac{1}{\lambda _{1}^{2}}+\frac{1}{\lambda _{2}^{2}}}} $
$ =\frac{2{\lambda_{1}}{\lambda_{2}}}{\sqrt{\lambda _{1}^{2}+\lambda _{2}^{2}}} $
BETA
