A) it is only associated with atomic particles
B) \[\frac{h}{2KEm}\]
C) \[\sqrt{\frac{h}{2KEm}}\]
D) \[\frac{h}{\sqrt{2KEm}}\]
Correct Answer: D
Solution :
de-Broglie wavelength \[(\lambda )\] is given by \[\lambda =\frac{h}{p}\] ?. (i) where h is Plancks constant, p the momentum. Kinetic energy \[KE=\frac{1}{2}m{{v}^{2}}=\frac{{{p}^{2}}}{2\,m}\] ... (ii) where p is momentum. From Eq. (ii) \[p=\sqrt{2\,m\,KE}\] \[\therefore \] \[\therefore \lambda =\frac{h}{\sqrt{2\,m\,KE}}\]
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