The de-broglie wavelength of 2 elementary particles in centre of mass frame is same and is equal to 5/8 times de-broglie wavelength of the lighter particle in lab frame and 5/2 times the de-broglie wavelength of the heavier particle in lab frame. Both particles have either parallel or anti parallel velocities. The possible combinations of particles are :

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$_{6}^{13} C a n d_{6}^{12} C$

$_{1}^{1} H a n d_{2}^{4} H e$

$_{6}^{13} C a n d_{2}^{4} H e$

$_{3}^{7} L i a n d_{6}^{13} C$

answer is D.

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Detailed Solution

$\frac{h}{μ v_{r e l}} = \frac{5}{8} \frac{h}{m_{1} v_{1}} = \frac{5}{8} \frac{h}{m_{2} v_{2}}$ [ $μ v_{r e l} =$ momentum of each particle in C – frame $v_{1}, v_{2}$ are velocities in ground frame and $m_{1}$ = mass of lighter particle]
$m_{2} v_{2} = \pm 4 m_{1} v_{1}$
Now four possibilities will arise for checking
Question Image

Case – 1 : $m_{2} v_{2} = + 4 m_{1} v_{1}, v_{r e l} = v_{1} - v_{2}$
$\frac{m_{1} m_{2}}{m_{1} + m_{2}} (v_{1} - v_{2}) = \frac{8}{5} m_{1} v_{1} m_{2} v_{1} - m_{2} v_{2} = \frac{8}{5} m_{1} v_{1} + \frac{8}{5} m_{2} v_{1} \frac{28}{5} m_{1} v_{1} = - \frac{3}{5} m_{2} v_{1}$
$\frac{m_{1}}{m_{2}} = - \frac{3}{28}$ not possible as $m_{1} > 0, m_{2} > 0$
Case – 2 : $v_{r e l} = v_{2} - v_{1}, m_{2} v_{2} = - 4 m_{1} v_{1}$
$\Rightarrow \frac{m_{2} (v_{2} - v_{1})}{m_{1} + m_{2}} = \frac{8}{5} v_{1} m_{2} v_{2} - m_{2} v_{1} = \frac{8}{5} m_{1} v_{1} + \frac{8}{5} m_{2} v_{1} \frac{3}{5} m_{2} v_{2} = \frac{13}{5} m_{2} v_{1} \Rightarrow v_{2} = \frac{13}{3} v_{1} m_{2} v_{2} = 4 m_{1} v_{1}$
$\frac{v_{2}}{v_{1}} = \frac{4 m_{1}}{m_{2}} = \frac{13}{3} \Rightarrow \frac{m_{1}}{m_{2}} = \frac{13}{12}$ not possible as $m_{1} < m_{2}$
Case – 3 : $m_{2} v_{2} = - 4 m_{1} v_{1} a n d v_{r e l} = v_{1} - v_{2}$
$m_{2} v_{1} - m_{2} v_{2} = + \frac{8}{5} m_{1} v_{1} + \frac{8}{5} m_{2} v_{1} \Rightarrow - \frac{3}{5} m_{2} v_{1} = \frac{3}{5} m_{2} v_{2}$
$v_{1} = - v_{2} \Rightarrow m_{2} (- v_{1}) = - 4 m_{1} v_{1} \Rightarrow \frac{m_{2}}{m_{1}} = 4.$ This is possible
Case – 4 : $m_{2} v_{2} = - 4 m_{1} v_{1} a n d v_{r e l} = v_{2} - v_{1}$
$m_{2} v_{2} - m_{2} v_{1} = \frac{8}{5} m_{1} v_{1} + \frac{8}{5} m_{2} v_{1} m_{2} v_{2} - \frac{8}{5} m_{1} v_{1} = \frac{13}{5} m_{2} v_{1} \Rightarrow \frac{7}{5} m_{2} v_{2} = \frac{13}{5} m_{2} v_{1}$
$\frac{v_{2}}{v_{1}} = \frac{13}{7} not possible as \frac{m_{2}}{m_{1}} will be - v e$