A) Both have a free fall through the same height
B) Both move with the same kinetic energy
C) Both move with the same linear momentum
D) Both move with the same speed
Correct Answer: A
Solution :
The de-Broglie equation is \[\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mk}}\Rightarrow m=\frac{h}{\lambda v}\] \[\Rightarrow \] \[m\propto \frac{1}{\lambda },\]if \[\frac{h}{v}=\operatorname{constant}\] Again, \[\frac{h}{v}\]will be constant only when v is constant. Thus, in present case the de-Broglie wavelength of heavier particle will be smaller. When the velocities of two particles will be same or both move with same kinetic energy (K).
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