A) A positron
B) A photon
C) An \[\alpha -\]particle
D) A neutron
Correct Answer: C
Solution :
The de-Broglie equation is\[\lambda =\frac{h}{p}=\frac{h}{mv}\]. Here, \[h\] and \[v\] are constant. So\[,\,\,\lambda \propto \frac{1}{m}\]. Since, the \[\alpha -\]particle has the highest mass among the given entities, it has the smallest de-Broglie wavelength.You need to login to perform this action.
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