A) zero
B) infinity
C) \[\frac{hv}{c}\]
D) \[\frac{{{m}_{0}}c}{h}\]
Correct Answer: A
Solution :
The de-Broglie wavelength \[(\lambda )\] is given by \[\lambda =\frac{h}{p}\] ...(i) where h is Planck's constant, p the momentum. Also, \[p=mv=\frac{{{m}_{0}}}{\sqrt{1-\frac{{{v}^{2}}}{{{C}^{2}}}}}\] ...(ii) where \[{{m}_{0}}\] is rest mass, v the velocity and c the speed of light. From Eqs. (i) and (ii), we get \[\lambda =\frac{h\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}}{{{m}_{0}}v}\] Given: \[v=c\] \[\therefore \] \[\lambda =0\]
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